What are the solutions to the equation quadraticformula 3x+18=x^2 ?

The solutions to the equation quadraticformula 3x+18=x^2 are x=6,x=-3

Find the zeros of quadraticformula 3x+18=x^2

The zeros of quadraticformula 3x+18=x^2 are x=6,x=-3

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3 x + 18 = x^{2}

x = 6, x = - 3

Solution steps

3 x + 18 = x^{2}

Switch sides

x^{2} = 3 x + 18

Move

18

to the left side

x^{2} - 18 = 3 x

Move

3 x

to the left side

x^{2} - 18 - 3 x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - 3 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 18 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 18) = 9

x_{1, 2} = \frac{- ( - 3 ) \pm 9}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 9}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 9}{2 \cdot 1}

x = \frac{- ( - 3 ) + 9}{2 \cdot 1} : 6

x = \frac{- ( - 3 ) - 9}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

x = 6, x = - 3

(6 - x)^{2} = 64

x, f (x + 3) = 3 x^{2} + 3 x - 3

(2 - x) (x + 1) - x (x - 1) = 2

18 = (- 6 + 3)^{2} + (2 - y)^{2}

(3 n - 2) = \frac{n ( 3 n - 1 )}{2}

What are the solutions to the equation quadraticformula 3x+18=x^2 ?
The solutions to the equation quadraticformula 3x+18=x^2 are x=6,x=-3
Find the zeros of quadraticformula 3x+18=x^2
The zeros of quadraticformula 3x+18=x^2 are x=6,x=-3