What are the solutions to the equation quadraticformula (x-2)(x+10)=13 ?

The solutions to the equation quadraticformula (x-2)(x+10)=13 are x=3,x=-11

Find the zeros of quadraticformula (x-2)(x+10)=13

The zeros of quadraticformula (x-2)(x+10)=13 are x=3,x=-11

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(x - 2) (x + 10) = 13

x = 3, x = - 11

Solution steps

(x - 2) (x + 10) = 13

Expand

(x - 2) (x + 10) : x^{2} + 8 x - 20

x^{2} + 8 x - 20 = 13

Move

13

to the left side

x^{2} + 8 x - 33 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot ( - 33 )}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot (- 33) = 14

x_{1, 2} = \frac{- 8 \pm 14}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 8 + 14}{2 \cdot 1}, x_{2} = \frac{- 8 - 14}{2 \cdot 1}

x = \frac{- 8 + 14}{2 \cdot 1} : 3

x = \frac{- 8 - 14}{2 \cdot 1} : - 11

The solutions to the quadratic equation are:

x = 3, x = - 11

x (x - 8) = 0

x, 25 6^{x} = 64

x, 2 x = x^{2}

R, R = \frac{( \frac{200}{3} )}{1.5} \cdot \frac{12000}{3}

r, - 10 - 5 r + 7 r = 10 - 8 (- 4 r - 5)

What are the solutions to the equation quadraticformula (x-2)(x+10)=13 ?
The solutions to the equation quadraticformula (x-2)(x+10)=13 are x=3,x=-11
Find the zeros of quadraticformula (x-2)(x+10)=13
The zeros of quadraticformula (x-2)(x+10)=13 are x=3,x=-11