3(x+4)^2+15(x+4)-18=0 — step-by-step solution

Let's solve in

ℝ

3 \cdot {(x + 4)}^{2} + 15 \cdot (x + 4) - 18 = 0

ⓘ Raising a sum to an integer power is multiplying it by itself:

{(x + 4)}^{2}

(x + 4) \cdot (x + 4)

■Expand by multiplying each term inside the parentheses:

15 \cdot (x + 4)

15 x + 60

3 \cdot (x + 4) \cdot (x + 4)

+ 15 x

+ 60

- 18 = 0

ⓘ Start from

60

and move back

18

on the number line:

60 - 18

42

3 \cdot (x + 4) \cdot (x + 4) + 15 x + 42 = 0

ⓘ Expand by multiplying each term inside the parentheses:

3 \cdot (x + 4)

3 x + 12

(3 x + 12) \cdot (x + 4) + 15 x + 42 = 0

ⓘ Expand by multiplying each term inside the parentheses:

(3 x + 12) \cdot (x + 4)

3 \cdot x \cdot x

+ 3 x \cdot 4

+ 12 x

+ 12 \cdot 4

3 \cdot x \cdot x

+ 3 x \cdot 4

+ 12 x

+ 12 \cdot 4

+ 15 x

+ 42 = 0

ⓘ Multiplying powers with the same base adds the exponents:

x \cdot x

x^{2}

■Multiplications are done before additions and subtractions:

3 \cdot 4

12

12 \cdot 4

(10 + 2) \cdot 4

10 \cdot 4 + 2 \cdot 4

40 + 8

48

■Like terms are grouped by adding their coefficients:

12 x + 15 x

27 x

3 x^{2}

+ 12 x

+ 27 x

+ 48

+ 42 = 0

ⓘ Like terms are grouped by adding their coefficients:

12 x + 27 x

39 x

■Start from

48

and move forward

42

on the number line:

48 + 42

90

3 x^{2} + 39 x + 90 = 0

Since

3 \neq 0

, divide both sides by

3

x^{2} + 13 x + 30 = 0

Given the quadratic form

a x^{2} + b x + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = 13

and

c = 30

Δ = 13^{2} - 4 \cdot 1 \cdot 30

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

13^{2}

13 \cdot 13

169

Δ = 169 - 4 \cdot 30

ⓘ Multiplications are done before additions and subtractions:

4 \cdot 30

120

Δ = 169 - 120

ⓘ Start from

169

and move back

120

on the number line:

169 - 120

49

Δ = 49

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 13 - \sqrt{49}}{2 \cdot 1}

x_{1} = \frac{- 13 - \sqrt{49}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{49}

7

;

7^{2}

49

x_{1} = \frac{- 13 - 7}{2}

ⓘ The numerator and the denominator are computed before the division:

- 13 - 7

-20

x_{1} = - \frac{20}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{20}{2}

-10

x_{1} = -10

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 13 + \sqrt{49}}{2 \cdot 1}

x_{2} = \frac{- 13 + \sqrt{49}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{49}

7

;

7^{2}

49

x_{2} = \frac{- 13 + 7}{2}

ⓘ The numerator and the denominator are computed before the division:

- 13 + 7

-6

x_{2} = - \frac{6}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{6}{2}

-3

x_{2} = -3

▶Solutions obtained:

x = {-10, -3}