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2x^2-9x-3=12-6x
We move all terms to the left:
2x^2-9x-3-(12-6x)=0
We add all the numbers together, and all the variables
2x^2-9x-(-6x+12)-3=0
We get rid of parentheses
2x^2-9x+6x-12-3=0
We add all the numbers together, and all the variables
2x^2-3x-15=0
a = 2; b = -3; c = -15;
Δ = b2-4ac
Δ = -32-4·2·(-15)
Δ = 129
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{-(-3)-\sqrt{129}}{2*2}=\frac{3-\sqrt{129}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{129}}{2*2}=\frac{3+\sqrt{129}}{4} $
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