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(X+4)(X-5)+X=180
We move all terms to the left:
(X+4)(X-5)+X-(180)=0
We add all the numbers together, and all the variables
X+(X+4)(X-5)-180=0
We multiply parentheses ..
(+X^2-5X+4X-20)+X-180=0
We get rid of parentheses
X^2-5X+4X+X-20-180=0
We add all the numbers together, and all the variables
X^2-200=0
a = 1; b = 0; c = -200;
Δ = b2-4ac
Δ = 02-4·1·(-200)
Δ = 800
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{2}}{2*1}=\frac{0-20\sqrt{2}}{2} =-\frac{20\sqrt{2}}{2} =-10\sqrt{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{2}}{2*1}=\frac{0+20\sqrt{2}}{2} =\frac{20\sqrt{2}}{2} =10\sqrt{2} $
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