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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^2-15x-15=0
a = 2; b = -15; c = -15;
Δ = b2-4ac
Δ = -152-4·2·(-15)
Δ = 345
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{-(-15)-\sqrt{345}}{2*2}=\frac{15-\sqrt{345}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{345}}{2*2}=\frac{15+\sqrt{345}}{4} $
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