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2x^2+20x=-44
We move all terms to the left:
2x^2+20x-(-44)=0
We add all the numbers together, and all the variables
2x^2+20x+44=0
a = 2; b = 20; c = +44;
Δ = b2-4ac
Δ = 202-4·2·44
Δ = 48
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{3}}{2*2}=\frac{-20-4\sqrt{3}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{3}}{2*2}=\frac{-20+4\sqrt{3}}{4} $
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