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(2x+10)x=170
We move all terms to the left:
(2x+10)x-(170)=0
We multiply parentheses
2x^2+10x-170=0
a = 2; b = 10; c = -170;
Δ = b2-4ac
Δ = 102-4·2·(-170)
Δ = 1460
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1460}=\sqrt{4*365}=\sqrt{4}*\sqrt{365}=2\sqrt{365}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{365}}{2*2}=\frac{-10-2\sqrt{365}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{365}}{2*2}=\frac{-10+2\sqrt{365}}{4} $
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