What are the solutions to the equation 8x^2+7=x^2-8x+8 ?

The solutions to the equation 8x^2+7=x^2-8x+8 are x=-(4+sqrt(23))/7 ,x=(sqrt(23)-4)/7

Find the zeros of 8x^2+7=x^2-8x+8

The zeros of 8x^2+7=x^2-8x+8 are x=-(4+sqrt(23))/7 ,x=(sqrt(23)-4)/7

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8 x^{2} + 7 = x^{2} - 8 x + 8

x = - \frac{4 + 23}{7}, x = \frac{23 - 4}{7}

Decimal Notation

x = - 1.25654 \dots, x = 0.11369 \dots

Solution steps

8 x^{2} + 7 = x^{2} - 8 x + 8

Switch sides

x^{2} - 8 x + 8 = 8 x^{2} + 7

Move

7

to the left side

x^{2} - 8 x + 1 = 8 x^{2}

Move

8 x^{2}

to the left side

- 7 x^{2} - 8 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 7 ) \cdot 1}{2 ( - 7 )}

(- 8)^{2} - 4 (- 7) \cdot 1 = 223

x_{1, 2} = \frac{- ( - 8 ) \pm 2 23}{2 ( - 7 )}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 2 23}{2 ( - 7 )}, x_{2} = \frac{- ( - 8 ) - 2 23}{2 ( - 7 )}

x = \frac{- ( - 8 ) + 2 23}{2 ( - 7 )} : - \frac{4 + 23}{7}

x = \frac{- ( - 8 ) - 2 23}{2 ( - 7 )} : \frac{23 - 4}{7}

The solutions to the quadratic equation are:

x = - \frac{4 + 23}{7}, x = \frac{23 - 4}{7}

x^{\frac{- 9}{2}} + 2 x^{- \frac{7}{2}} + x^{\frac{- 5}{2}}

Y = (0)^{2}

\frac{- 6}{- 7}

e^{i \cdot \frac{π}{2}}

r^{3} - 125

What are the solutions to the equation 8x^2+7=x^2-8x+8 ?
The solutions to the equation 8x^2+7=x^2-8x+8 are x=-(4+sqrt(23))/7 ,x=(sqrt(23)-4)/7
Find the zeros of 8x^2+7=x^2-8x+8
The zeros of 8x^2+7=x^2-8x+8 are x=-(4+sqrt(23))/7 ,x=(sqrt(23)-4)/7