y^2+3y-6=0 — step-by-step solution

1) Let's solve in

ℝ

y^{2} + 3 y - 6 = 0

Given the quadratic form

a y^{2} + b y + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = 3

and

c = -6

Δ = 3^{2} - 4 \cdot 1 \cdot (-6)

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

3^{2}

3 \cdot 3

9

Δ = 9 - 4 \cdot (-6)

ⓘ Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive:

- 4 \cdot (-6)

24

Δ = 9 + 24

ⓘ Start from

9

and move forward

24

on the number line:

9 + 24

33

Δ = 33

Δ > 0

, the equation has two roots:

The quadratic formula gives

y_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 3 - \sqrt{33}}{2 \cdot 1}

y_{1} = \frac{- 3 - \sqrt{33}}{2}

ⓘ The minus sign is factored out, changing the sign of each term:

- 3 - \sqrt{33}

- (3 + \sqrt{33})

y_{1} = - \frac{3 + \sqrt{33}}{2}

The quadratic formula gives

y_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 3 + \sqrt{33}}{2 \cdot 1}

y_{2} = \frac{- 3 + \sqrt{33}}{2}

▶Solutions obtained:

y = {- \frac{3 + \sqrt{33}}{2}, \frac{\sqrt{33} - 3}{2}}

2) The numeric value of:

y = {- \frac{3 + \sqrt{33}}{2}, \frac{\sqrt{33} - 3}{2}}

equals:

y = {\approx - \frac{3 + 5.7446}{2}, \approx \frac{5.7446 - 3}{2}}

y = {\approx - \frac{8.7446}{2}, \approx \frac{2.7446}{2}}

▷

y = {\approx -4.3723, \approx 1.3723}