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(x-4)(x-4)=3x-12
We move all terms to the left:
(x-4)(x-4)-(3x-12)=0
We get rid of parentheses
(x-4)(x-4)-3x+12=0
We multiply parentheses ..
(+x^2-4x-4x+16)-3x+12=0
We get rid of parentheses
x^2-4x-4x-3x+16+12=0
We add all the numbers together, and all the variables
x^2-11x+28=0
a = 1; b = -11; c = +28;
Δ = b2-4ac
Δ = -112-4·1·28
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-3}{2*1}=\frac{8}{2} =4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+3}{2*1}=\frac{14}{2} =7 $
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