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(x-3)(x-4)=11x+44
We move all terms to the left:
(x-3)(x-4)-(11x+44)=0
We get rid of parentheses
(x-3)(x-4)-11x-44=0
We multiply parentheses ..
(+x^2-4x-3x+12)-11x-44=0
We get rid of parentheses
x^2-4x-3x-11x+12-44=0
We add all the numbers together, and all the variables
x^2-18x-32=0
a = 1; b = -18; c = -32;
Δ = b2-4ac
Δ = -182-4·1·(-32)
Δ = 452
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{452}=\sqrt{4*113}=\sqrt{4}*\sqrt{113}=2\sqrt{113}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{113}}{2*1}=\frac{18-2\sqrt{113}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{113}}{2*1}=\frac{18+2\sqrt{113}}{2} $
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