What are the solutions to the equation 8x^2+7x-15=-7 ?

The solutions to the equation 8x^2+7x-15=-7 are x=(-7+sqrt(305))/(16),x=(-7-sqrt(305))/(16)

Find the zeros of 8x^2+7x-15=-7

The zeros of 8x^2+7x-15=-7 are x=(-7+sqrt(305))/(16),x=(-7-sqrt(305))/(16)

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8 x^{2} + 7 x - 15 = - 7

x = \frac{- 7 + 305}{16}, x = \frac{- 7 - 305}{16}

Decimal Notation

x = 0.65401 \dots, x = - 1.52901 \dots

Solution steps

8 x^{2} + 7 x - 15 = - 7

Move

7

to the left side

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

Solve with the quadratic formula

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

7^{2} - 4 \cdot 8 (- 8) = 305

x_{1, 2} = \frac{- 7 \pm 305}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- 7 + 305}{2 \cdot 8}, x_{2} = \frac{- 7 - 305}{2 \cdot 8}

x = \frac{- 7 + 305}{2 \cdot 8} : \frac{- 7 + 305}{16}

x = \frac{- 7 - 305}{2 \cdot 8} : \frac{- 7 - 305}{16}

The solutions to the quadratic equation are:

x = \frac{- 7 + 305}{16}, x = \frac{- 7 - 305}{16}

\frac{1}{a} + \frac{1}{b}

ln (x) < 0

- 2 ∣ x - 14 ∣ + 5 = - 6 ∣ x - 14 ∣ - 1

x^{4} - 16 y^{4}

7 + 2 (3 x + 4) = 3

What are the solutions to the equation 8x^2+7x-15=-7 ?
The solutions to the equation 8x^2+7x-15=-7 are x=(-7+sqrt(305))/(16),x=(-7-sqrt(305))/(16)
Find the zeros of 8x^2+7x-15=-7
The zeros of 8x^2+7x-15=-7 are x=(-7+sqrt(305))/(16),x=(-7-sqrt(305))/(16)