What are the solutions to the equation 8x^2+52x-28=0 ?

The solutions to the equation 8x^2+52x-28=0 are x= 1/2 ,x=-7

Find the zeros of 8x^2+52x-28=0

The zeros of 8x^2+52x-28=0 are x= 1/2 ,x=-7

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8 x^{2} + 52 x - 28 = 0

x = \frac{1}{2}, x = - 7

Decimal Notation

x = 0.5, x = - 7

Solution steps

8 x^{2} + 52 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 52 \pm 5 2 ^{2} - 4 \cdot 8 ( - 28 )}{2 \cdot 8}

5 2^{2} - 4 \cdot 8 (- 28) = 60

x_{1, 2} = \frac{- 52 \pm 60}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- 52 + 60}{2 \cdot 8}, x_{2} = \frac{- 52 - 60}{2 \cdot 8}

x = \frac{- 52 + 60}{2 \cdot 8} : \frac{1}{2}

x = \frac{- 52 - 60}{2 \cdot 8} : - 7

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = - 7

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

\frac{4}{7} y = 8

\frac{x + 2}{7} = \frac{5 x + 7}{9}

n^{2} - 3 n + 2 = 0

18 y^{3} x^{2} - 2 x y^{4}

What are the solutions to the equation 8x^2+52x-28=0 ?
The solutions to the equation 8x^2+52x-28=0 are x= 1/2 ,x=-7
Find the zeros of 8x^2+52x-28=0
The zeros of 8x^2+52x-28=0 are x= 1/2 ,x=-7