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x2+2(2x-5)3=x+156
We move all terms to the left:
x2+2(2x-5)3-(x+156)=0
We add all the numbers together, and all the variables
x^2+2(2x-5)3-(x+156)=0
We multiply parentheses
x^2+12x-(x+156)-30=0
We get rid of parentheses
x^2+12x-x-156-30=0
We add all the numbers together, and all the variables
x^2+11x-186=0
a = 1; b = 11; c = -186;
Δ = b2-4ac
Δ = 112-4·1·(-186)
Δ = 865
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{865}}{2*1}=\frac{-11-\sqrt{865}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{865}}{2*1}=\frac{-11+\sqrt{865}}{2} $
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