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-3x^2-0x+36=0
We add all the numbers together, and all the variables
-3x^2-1x+36=0
a = -3; b = -1; c = +36;
Δ = b2-4ac
Δ = -12-4·(-3)·36
Δ = 433
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{-(-1)-\sqrt{433}}{2*-3}=\frac{1-\sqrt{433}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{433}}{2*-3}=\frac{1+\sqrt{433}}{-6} $
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