What are the solutions to the equation q^2-4q=12 ?

The solutions to the equation q^2-4q=12 are q=6,q=-2

Find the zeros of q^2-4q=12

The zeros of q^2-4q=12 are q=6,q=-2

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q^{2} - 4 q = 12

q = 6, q = - 2

Solution steps

q^{2} - 4 q = 12

Move

12

to the left side

q^{2} - 4 q - 12 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 12) = 8

q_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 \cdot 1}

Separate the solutions

q_{1} = \frac{- ( - 4 ) + 8}{2 \cdot 1}, q_{2} = \frac{- ( - 4 ) - 8}{2 \cdot 1}

q = \frac{- ( - 4 ) + 8}{2 \cdot 1} : 6

q = \frac{- ( - 4 ) - 8}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

q = 6, q = - 2

4 z^{2} + 11 z = 3

- 4 + 4 x - x^{2} = 0

x, x^{2} - x - 72 = 0

5 x + 8 = 2 x^{2}

4 x = 4 - x^{2}

What are the solutions to the equation q^2-4q=12 ?
The solutions to the equation q^2-4q=12 are q=6,q=-2
Find the zeros of q^2-4q=12
The zeros of q^2-4q=12 are q=6,q=-2