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(x+5)(x-7)=28
We move all terms to the left:
(x+5)(x-7)-(28)=0
We multiply parentheses ..
(+x^2-7x+5x-35)-28=0
We get rid of parentheses
x^2-7x+5x-35-28=0
We add all the numbers together, and all the variables
x^2-2x-63=0
a = 1; b = -2; c = -63;
Δ = b2-4ac
Δ = -22-4·1·(-63)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-16}{2*1}=\frac{-14}{2} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+16}{2*1}=\frac{18}{2} =9 $
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