What are the solutions to the equation 7x-8=-3x^2 ?

The solutions to the equation 7x-8=-3x^2 are x=-(7+sqrt(145))/6 ,x=(sqrt(145)-7)/6

Find the zeros of 7x-8=-3x^2

The zeros of 7x-8=-3x^2 are x=-(7+sqrt(145))/6 ,x=(sqrt(145)-7)/6

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7 x - 8 = - 3 x^{2}

x = - \frac{7 + 145}{6}, x = \frac{145 - 7}{6}

Solution steps

7 x - 8 = - 3 x^{2}

Switch sides

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

Move

8

to the left side

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

Move

7 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 3 ) \cdot 8}{2 ( - 3 )}

(- 7)^{2} - 4 (- 3) \cdot 8 = 145

x_{1, 2} = \frac{- ( - 7 ) \pm 145}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 145}{2 ( - 3 )}, x_{2} = \frac{- ( - 7 ) - 145}{2 ( - 3 )}

x = \frac{- ( - 7 ) + 145}{2 ( - 3 )} : - \frac{7 + 145}{6}

x = \frac{- ( - 7 ) - 145}{2 ( - 3 )} : \frac{145 - 7}{6}

The solutions to the quadratic equation are:

x = - \frac{7 + 145}{6}, x = \frac{145 - 7}{6}

15 x^{2} - x - 6 = 0

- 3 (3 x + 15) - (10 + x) = 35

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

7 t - 6 v^{2} + t v - 42 v

16 m^{2} - 40 mn + 25 n^{2}

What are the solutions to the equation 7x-8=-3x^2 ?
The solutions to the equation 7x-8=-3x^2 are x=-(7+sqrt(145))/6 ,x=(sqrt(145)-7)/6
Find the zeros of 7x-8=-3x^2
The zeros of 7x-8=-3x^2 are x=-(7+sqrt(145))/6 ,x=(sqrt(145)-7)/6