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(2x+10)(x-10)=180
We move all terms to the left:
(2x+10)(x-10)-(180)=0
We multiply parentheses ..
(+2x^2-20x+10x-100)-180=0
We get rid of parentheses
2x^2-20x+10x-100-180=0
We add all the numbers together, and all the variables
2x^2-10x-280=0
a = 2; b = -10; c = -280;
Δ = b2-4ac
Δ = -102-4·2·(-280)
Δ = 2340
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2340}=\sqrt{36*65}=\sqrt{36}*\sqrt{65}=6\sqrt{65}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-6\sqrt{65}}{2*2}=\frac{10-6\sqrt{65}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+6\sqrt{65}}{2*2}=\frac{10+6\sqrt{65}}{4} $
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