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(X-4)(X-12)=105
We move all terms to the left:
(X-4)(X-12)-(105)=0
We multiply parentheses ..
(+X^2-12X-4X+48)-105=0
We get rid of parentheses
X^2-12X-4X+48-105=0
We add all the numbers together, and all the variables
X^2-16X-57=0
a = 1; b = -16; c = -57;
Δ = b2-4ac
Δ = -162-4·1·(-57)
Δ = 484
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{484}=22$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-22}{2*1}=\frac{-6}{2} =-3 $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+22}{2*1}=\frac{38}{2} =19 $
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