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135=(x+1)(x-5)
We move all terms to the left:
135-((x+1)(x-5))=0
We multiply parentheses ..
-((+x^2-5x+x-5))+135=0
We calculate terms in parentheses: -((+x^2-5x+x-5)), so:We get rid of parentheses
(+x^2-5x+x-5)
We get rid of parentheses
x^2-5x+x-5
We add all the numbers together, and all the variables
x^2-4x-5
Back to the equation:
-(x^2-4x-5)
-x^2+4x+5+135=0
We add all the numbers together, and all the variables
-1x^2+4x+140=0
a = -1; b = 4; c = +140;
Δ = b2-4ac
Δ = 42-4·(-1)·140
Δ = 576
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{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-24}{2*-1}=\frac{-28}{-2} =+14 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+24}{2*-1}=\frac{20}{-2} =-10 $
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