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(X+4)(X-5=-8
We move all terms to the left:
(X+4)(X-5-(-8)=0
We calculate terms in parentheses: +(X+4)(X-5-(-8), so:We multiply parentheses
X+4)(X-5-(-8
We add all the numbers together, and all the variables
X+4)(X
Back to the equation:
+(X+4)(X)
X^2+4X=0
a = 1; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·1·0
Δ = 16
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{16}=4$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*1}=\frac{-8}{2} =-4 $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*1}=\frac{0}{2} =0 $
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