5x^2-13x=6 — step-by-step solution

1) Let's solve in

ℝ

5 x^{2} - 13 x = 6

5 x^{2} - 13 x - 6 = 0

Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

Δ = {(-13)}^{2} - 4 \cdot 5 \cdot (-6)

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

{(-13)}^{2}

(-13) \cdot (-13)

169

■Multiplications are done before additions and subtractions:

4 \cdot 5

20

■Two minus signs cancel out, the product of two negative numbers is positive:

- 20 \cdot (-6)

20 \cdot 6

Δ = 169 + 20 \cdot 6

ⒾMultiplications are done before additions and subtractions:

20 \cdot 6

120

Δ = 169 + 120

ⒾStart from

169

and move forward

120

on the number line:

169 + 120

289

Δ = 289

Since

Δ > 0

, the equation has two roots:

For the first root, we obtain:

\frac{- b - \sqrt{Δ}}{2 a}

x1 = \frac{13 - \sqrt{289}}{2 \cdot 5}

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

- \sqrt{289}

-17

;

17^{2}

289

ℹ

2 \cdot 5

10

x1 = \frac{13 - 17}{10}

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

13 - 17

-4

x1 = - \frac{4}{10}

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

2

, their greatest common divisor:

- \frac{4}{10}

- \frac{2 \cdot 2}{5 \cdot 2}

- \frac{2}{5}

x1 = - \frac{2}{5}

For the second root, we obtain:

\frac{- b + \sqrt{Δ}}{2 a}

x2 = \frac{13 + \sqrt{289}}{2 \cdot 5}

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

\sqrt{289}

17

;

17^{2}

289

ℹ

2 \cdot 5

10

x2 = \frac{13 + 17}{10}

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

13 + 17

30

x2 = \frac{30}{10}

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

10

, their greatest common divisor:

\frac{30}{10}

3

x2 = 3

▶Solutions obtained:

x = {- \frac{2}{5}, 3}

2) The numeric value of:

x = {- \frac{2}{5}, 3}

equals:

▷

x = {-0.4, 3}