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(x-4)(x+3)-18=0
We multiply parentheses ..
(+x^2+3x-4x-12)-18=0
We get rid of parentheses
x^2+3x-4x-12-18=0
We add all the numbers together, and all the variables
x^2-1x-30=0
a = 1; b = -1; c = -30;
Δ = b2-4ac
Δ = -12-4·1·(-30)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-11}{2*1}=\frac{-10}{2} =-5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+11}{2*1}=\frac{12}{2} =6 $
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