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2x^2+12x=3x-5
We move all terms to the left:
2x^2+12x-(3x-5)=0
We get rid of parentheses
2x^2+12x-3x+5=0
We add all the numbers together, and all the variables
2x^2+9x+5=0
a = 2; b = 9; c = +5;
Δ = b2-4ac
Δ = 92-4·2·5
Δ = 41
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{-(9)-\sqrt{41}}{2*2}=\frac{-9-\sqrt{41}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{41}}{2*2}=\frac{-9+\sqrt{41}}{4} $
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