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(2x-11)(2x-2)=180
We move all terms to the left:
(2x-11)(2x-2)-(180)=0
We multiply parentheses ..
(+4x^2-4x-22x+22)-180=0
We get rid of parentheses
4x^2-4x-22x+22-180=0
We add all the numbers together, and all the variables
4x^2-26x-158=0
a = 4; b = -26; c = -158;
Δ = b2-4ac
Δ = -262-4·4·(-158)
Δ = 3204
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3204}=\sqrt{36*89}=\sqrt{36}*\sqrt{89}=6\sqrt{89}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-6\sqrt{89}}{2*4}=\frac{26-6\sqrt{89}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+6\sqrt{89}}{2*4}=\frac{26+6\sqrt{89}}{8} $
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