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(2x+4)(x-3)=-12
We move all terms to the left:
(2x+4)(x-3)-(-12)=0
We add all the numbers together, and all the variables
(2x+4)(x-3)+12=0
We multiply parentheses ..
(+2x^2-6x+4x-12)+12=0
We get rid of parentheses
2x^2-6x+4x-12+12=0
We add all the numbers together, and all the variables
2x^2-2x=0
a = 2; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·2·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*2}=\frac{0}{4} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*2}=\frac{4}{4} =1 $
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