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(x-3)(x-7)=36
We move all terms to the left:
(x-3)(x-7)-(36)=0
We multiply parentheses ..
(+x^2-7x-3x+21)-36=0
We get rid of parentheses
x^2-7x-3x+21-36=0
We add all the numbers together, and all the variables
x^2-10x-15=0
a = 1; b = -10; c = -15;
Δ = b2-4ac
Δ = -102-4·1·(-15)
Δ = 160
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4\sqrt{10}}{2*1}=\frac{10-4\sqrt{10}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4\sqrt{10}}{2*1}=\frac{10+4\sqrt{10}}{2} $
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