7y=3y^2+2 — step-by-step solution

1) Let's solve in

ℝ

7 y = 3 y^{2} + 2

- 3 y^{2} + 7 y - 2 = 0

Since

-1 \neq 0

, divide both sides by

-1

3 y^{2} - 7 y + 2 = 0

Given the quadratic form

a y^{2} + b y + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 3

b = -7

and

c = 2

Δ = {(-7)}^{2} - 4 \cdot 3 \cdot 2

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

{(-7)}^{2}

(-7) \cdot (-7)

49

■Multiplications are done before additions and subtractions:

4 \cdot 3

12

Δ = 49 - 12 \cdot 2

ⓘ Multiplications are done before additions and subtractions:

12 \cdot 2

(10 + 2) \cdot 2

10 \cdot 2 + 2 \cdot 2

20 + 4

24

Δ = 49 - 24

ⓘ Start from

49

and move back

24

on the number line:

49 - 24

25

Δ = 25

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

y_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-7) - \sqrt{25}}{2 \cdot 3}

y_{1} = \frac{7 - \sqrt{25}}{2 \cdot 3}

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

\sqrt{25}

5

;

5^{2}

25

ⅰ

2 \cdot 3

6

y_{1} = \frac{7 - 5}{6}

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

7 - 5

2

y_{1} = \frac{2}{6}

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

2

, their greatest common divisor:

\frac{2}{6}

\frac{1 \cdot 2}{3 \cdot 2}

\frac{1}{3}

y_{1} = \frac{1}{3}

The quadratic formula gives

y_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-7) + \sqrt{25}}{2 \cdot 3}

y_{2} = \frac{7 + \sqrt{25}}{2 \cdot 3}

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

\sqrt{25}

5

;

5^{2}

25

ⅰ

2 \cdot 3

6

y_{2} = \frac{7 + 5}{6}

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

7 + 5

12

y_{2} = \frac{12}{6}

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

6

, their greatest common divisor:

\frac{12}{6}

2

y_{2} = 2

▶Solutions obtained:

y = {\frac{1}{3}, 2}

2) The numeric value of:

y = {\frac{1}{3}, 2}

equals:

▷

y = {\approx 0.3333, 2}