What are the solutions to the equation 2x^2-8x-10=32 ?

The solutions to the equation 2x^2-8x-10=32 are x=7,x=-3

Find the zeros of 2x^2-8x-10=32

The zeros of 2x^2-8x-10=32 are x=7,x=-3

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2 x^{2} - 8 x - 10 = 32

x = 7, x = - 3

Solution steps

2 x^{2} - 8 x - 10 = 32

Move

32

to the left side

2 x^{2} - 8 x - 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 ( - 42 )}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 (- 42) = 20

x_{1, 2} = \frac{- ( - 8 ) \pm 20}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 20}{2 \cdot 2}, x_{2} = \frac{- ( - 8 ) - 20}{2 \cdot 2}

x = \frac{- ( - 8 ) + 20}{2 \cdot 2} : 7

x = \frac{- ( - 8 ) - 20}{2 \cdot 2} : - 3

The solutions to the quadratic equation are:

x = 7, x = - 3

(9 + m) (- m + 9) = (9 + m) (- m + 9)

12 x - 4 - 9 x^{2} = 0

\frac{1}{6} + \frac{3}{8} - \frac{5}{12}

5 x + 3 = 2 x - 6

\frac{( x ^{2} - 5 x + 6 )}{( x ^{2} - 7 x + 12 )}

What are the solutions to the equation 2x^2-8x-10=32 ?
The solutions to the equation 2x^2-8x-10=32 are x=7,x=-3
Find the zeros of 2x^2-8x-10=32
The zeros of 2x^2-8x-10=32 are x=7,x=-3