What are the solutions to the equation quadraticformula (x-2)(x+5)=18 ?

The solutions to the equation quadraticformula (x-2)(x+5)=18 are x=4,x=-7

Find the zeros of quadraticformula (x-2)(x+5)=18

The zeros of quadraticformula (x-2)(x+5)=18 are x=4,x=-7

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(x - 2) (x + 5) = 18

x = 4, x = - 7

Solution steps

(x - 2) (x + 5) = 18

Expand

(x - 2) (x + 5) : x^{2} + 3 x - 10

x^{2} + 3 x - 10 = 18

Move

18

to the left side

x^{2} + 3 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 28) = 11

x_{1, 2} = \frac{- 3 \pm 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 11}{2 \cdot 1}, x_{2} = \frac{- 3 - 11}{2 \cdot 1}

x = \frac{- 3 + 11}{2 \cdot 1} : 4

x = \frac{- 3 - 11}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

x = 4, x = - 7

x^{2} - 6 x = 40

- 37 - 45

12 \frac{9}{10}

\frac{( x ^{4} + 36 )}{( x ^{2} - 8 )}

0.8%

What are the solutions to the equation quadraticformula (x-2)(x+5)=18 ?
The solutions to the equation quadraticformula (x-2)(x+5)=18 are x=4,x=-7
Find the zeros of quadraticformula (x-2)(x+5)=18
The zeros of quadraticformula (x-2)(x+5)=18 are x=4,x=-7