What are the solutions to the equation x^2+12x-35=10x ?

The solutions to the equation x^2+12x-35=10x are x=5,x=-7

Find the zeros of x^2+12x-35=10x

The zeros of x^2+12x-35=10x are x=5,x=-7

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x^{2} + 12 x - 35 = 10 x

x = 5, x = - 7

Solution steps

x^{2} + 12 x - 35 = 10 x

Move

10 x

to the left side

x^{2} + 2 x - 35 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 35 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 35) = 12

x_{1, 2} = \frac{- 2 \pm 12}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 12}{2 \cdot 1}, x_{2} = \frac{- 2 - 12}{2 \cdot 1}

x = \frac{- 2 + 12}{2 \cdot 1} : 5

x = \frac{- 2 - 12}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

x = 5, x = - 7

90 = 24 x - 0.8 x^{2}

(x + 2) (x + 3) = x^{2} - 10

\frac{6 y ^{3} z}{2 y z ^{2}}

5 x^{2} - x + 1 \leq 4 x^{2} - 3 x + 4

9 k^{2} + 24 k + 18

What are the solutions to the equation x^2+12x-35=10x ?
The solutions to the equation x^2+12x-35=10x are x=5,x=-7
Find the zeros of x^2+12x-35=10x
The zeros of x^2+12x-35=10x are x=5,x=-7