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2x^2-12x-26=10
We move all terms to the left:
2x^2-12x-26-(10)=0
We add all the numbers together, and all the variables
2x^2-12x-36=0
a = 2; b = -12; c = -36;
Δ = b2-4ac
Δ = -122-4·2·(-36)
Δ = 432
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12\sqrt{3}}{2*2}=\frac{12-12\sqrt{3}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12\sqrt{3}}{2*2}=\frac{12+12\sqrt{3}}{4} $
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