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X2+20X=10
We move all terms to the left:
X2+20X-(10)=0
We add all the numbers together, and all the variables
X^2+20X-10=0
a = 1; b = 20; c = -10;
Δ = b2-4ac
Δ = 202-4·1·(-10)
Δ = 440
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{440}=\sqrt{4*110}=\sqrt{4}*\sqrt{110}=2\sqrt{110}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{110}}{2*1}=\frac{-20-2\sqrt{110}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{110}}{2*1}=\frac{-20+2\sqrt{110}}{2} $
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