x^2=5x+2 — step-by-step solution

1) Let's solve in

ℝ

x^{2} = 5 x + 2

x^{2} - 5 x - 2 = 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 = -5

and

c = -2

Δ = {(-5)}^{2} - 4 \cdot 1 \cdot (-2)

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

{(-5)}^{2}

(-5) \cdot (-5)

25

Δ = 25 - 4 \cdot (-2)

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

- 4 \cdot (-2)

8

Δ = 25 + 8

ⓘ Start from

25

and move forward

8

on the number line:

25 + 8

33

Δ = 33

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-5) - \sqrt{33}}{2 \cdot 1}

x_{1} = \frac{5 - \sqrt{33}}{2}

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-5) + \sqrt{33}}{2 \cdot 1}

x_{2} = \frac{5 + \sqrt{33}}{2}

▶Solutions found:

x = {\frac{5 - \sqrt{33}}{2}, \frac{5 + \sqrt{33}}{2}}

2) The numeric value of:

x = {\frac{5 - \sqrt{33}}{2}, \frac{5 + \sqrt{33}}{2}}

is:

x = {\approx \frac{5 - 5.7446}{2}, \approx \frac{5 + 5.7446}{2}}

x = {\approx - \frac{0.7446}{2}, \approx \frac{10.7446}{2}}

▷

x = {\approx -0.3723, \approx 5.3723}