What are the solutions to the equation (x^2-2)=(4x+10) ?

The solutions to the equation (x^2-2)=(4x+10) are x=6,x=-2

Find the zeros of (x^2-2)=(4x+10)

The zeros of (x^2-2)=(4x+10) are x=6,x=-2

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(x^{2} - 2) = (4 x + 10)

x = 6, x = - 2

Solution steps

(x^{2} - 2) = (4 x + 10)

Refine

x^{2} - 2 = 4 x + 10

Move

10

to the left side

x^{2} - 12 = 4 x

Move

4 x

to the left side

x^{2} - 12 - 4 x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - 4 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 12) = 8

x_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 8}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 8}{2 \cdot 1}

x = \frac{- ( - 4 ) + 8}{2 \cdot 1} : 6

x = \frac{- ( - 4 ) - 8}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 6, x = - 2

2 (- \frac{1}{2})

216 x^{6} - 125 y^{6}

847

\frac{2}{a ^{2}} - \frac{3}{ab} + \frac{4}{b ^{2}}

2 x + 3 = 5 x + 24

What are the solutions to the equation (x^2-2)=(4x+10) ?
The solutions to the equation (x^2-2)=(4x+10) are x=6,x=-2
Find the zeros of (x^2-2)=(4x+10)
The zeros of (x^2-2)=(4x+10) are x=6,x=-2