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2x^2-40x=-172
We move all terms to the left:
2x^2-40x-(-172)=0
We add all the numbers together, and all the variables
2x^2-40x+172=0
a = 2; b = -40; c = +172;
Δ = b2-4ac
Δ = -402-4·2·172
Δ = 224
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{14}}{2*2}=\frac{40-4\sqrt{14}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{14}}{2*2}=\frac{40+4\sqrt{14}}{4} $
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