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2x^2+36x=-110
We move all terms to the left:
2x^2+36x-(-110)=0
We add all the numbers together, and all the variables
2x^2+36x+110=0
a = 2; b = 36; c = +110;
Δ = b2-4ac
Δ = 362-4·2·110
Δ = 416
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{26}}{2*2}=\frac{-36-4\sqrt{26}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{26}}{2*2}=\frac{-36+4\sqrt{26}}{4} $
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