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2x^2+10=20x+20
We move all terms to the left:
2x^2+10-(20x+20)=0
We get rid of parentheses
2x^2-20x-20+10=0
We add all the numbers together, and all the variables
2x^2-20x-10=0
a = 2; b = -20; c = -10;
Δ = b2-4ac
Δ = -202-4·2·(-10)
Δ = 480
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{30}}{2*2}=\frac{20-4\sqrt{30}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{30}}{2*2}=\frac{20+4\sqrt{30}}{4} $
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