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(X+4)(X-3)=-12
We move all terms to the left:
(X+4)(X-3)-(-12)=0
We add all the numbers together, and all the variables
(X+4)(X-3)+12=0
We multiply parentheses ..
(+X^2-3X+4X-12)+12=0
We get rid of parentheses
X^2-3X+4X-12+12=0
We add all the numbers together, and all the variables
X^2+X=0
a = 1; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·1·0
Δ = 1
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{1}=1$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*1}=\frac{-2}{2} =-1 $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*1}=\frac{0}{2} =0 $
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