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(x-3)(x+5)=-x^2-3x-3
We move all terms to the left:
(x-3)(x+5)-(-x^2-3x-3)=0
We get rid of parentheses
x^2+3x+(x-3)(x+5)+3=0
We multiply parentheses ..
x^2+(+x^2+5x-3x-15)+3x+3=0
We get rid of parentheses
x^2+x^2+5x-3x+3x-15+3=0
We add all the numbers together, and all the variables
2x^2+5x-12=0
a = 2; b = 5; c = -12;
Δ = b2-4ac
Δ = 52-4·2·(-12)
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*2}=\frac{-16}{4} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*2}=\frac{6}{4} =1+1/2 $
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