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(x-13)(x+3)=0
We multiply parentheses ..
(+x^2+3x-13x-39)=0
We get rid of parentheses
x^2+3x-13x-39=0
We add all the numbers together, and all the variables
x^2-10x-39=0
a = 1; b = -10; c = -39;
Δ = b2-4ac
Δ = -102-4·1·(-39)
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-16}{2*1}=\frac{-6}{2} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+16}{2*1}=\frac{26}{2} =13 $
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