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(x+5)(x-4)=-19
We move all terms to the left:
(x+5)(x-4)-(-19)=0
We add all the numbers together, and all the variables
(x+5)(x-4)+19=0
We multiply parentheses ..
(+x^2-4x+5x-20)+19=0
We get rid of parentheses
x^2-4x+5x-20+19=0
We add all the numbers together, and all the variables
x^2+x-1=0
a = 1; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·1·(-1)
Δ = 5
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{5}}{2*1}=\frac{-1-\sqrt{5}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{5}}{2*1}=\frac{-1+\sqrt{5}}{2} $
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