What are the solutions to the equation 4x-7=-5x^2 ?

The solutions to the equation 4x-7=-5x^2 are x=-(2+sqrt(39))/5 ,x=(sqrt(39)-2)/5

Find the zeros of 4x-7=-5x^2

The zeros of 4x-7=-5x^2 are x=-(2+sqrt(39))/5 ,x=(sqrt(39)-2)/5

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4 x - 7 = - 5 x^{2}

x = - \frac{2 + 39}{5}, x = \frac{39 - 2}{5}

Decimal Notation

x = - 1.64899 \dots, x = 0.84899 \dots

Solution steps

4 x - 7 = - 5 x^{2}

Switch sides

- 5 x^{2} = 4 x - 7

Move

7

to the left side

- 5 x^{2} + 7 = 4 x

Move

4 x

to the left side

- 5 x^{2} + 7 - 4 x = 0

Write in the standard form

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

- 5 x^{2} - 4 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 5 ) \cdot 7}{2 ( - 5 )}

(- 4)^{2} - 4 (- 5) \cdot 7 = 239

x_{1, 2} = \frac{- ( - 4 ) \pm 2 39}{2 ( - 5 )}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 39}{2 ( - 5 )}, x_{2} = \frac{- ( - 4 ) - 2 39}{2 ( - 5 )}

x = \frac{- ( - 4 ) + 2 39}{2 ( - 5 )} : - \frac{2 + 39}{5}

x = \frac{- ( - 4 ) - 2 39}{2 ( - 5 )} : \frac{39 - 2}{5}

The solutions to the quadratic equation are:

x = - \frac{2 + 39}{5}, x = \frac{39 - 2}{5}

\frac{11}{64} \cdot \frac{180}{π}

∣ 5 x - 4 ∣ + 4 = 7

5 x = - 4 x + 36

u^{2} - 4 u - 5 = 0

a \frac{1}{2} x a^{2}

What are the solutions to the equation 4x-7=-5x^2 ?
The solutions to the equation 4x-7=-5x^2 are x=-(2+sqrt(39))/5 ,x=(sqrt(39)-2)/5
Find the zeros of 4x-7=-5x^2
The zeros of 4x-7=-5x^2 are x=-(2+sqrt(39))/5 ,x=(sqrt(39)-2)/5