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

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

Find the zeros of quadraticformula x^2-12x=-35

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

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x^{2} - 12 x = - 35

x = 7, x = 5

Solution steps

x^{2} - 12 x = - 35

Move

35

to the left side

x^{2} - 12 x + 35 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 35}{2 \cdot 1}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 7, x = 5

4 \frac{1}{2} \cdot 2 \frac{1}{4}

x, 7 - 2 x = - 3

7 x - x

6400

x, 9 x - 3 = x + 14

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