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0=6x^2+3x-33
We move all terms to the left:
0-(6x^2+3x-33)=0
We add all the numbers together, and all the variables
-(6x^2+3x-33)=0
We get rid of parentheses
-6x^2-3x+33=0
a = -6; b = -3; c = +33;
Δ = b2-4ac
Δ = -32-4·(-6)·33
Δ = 801
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{801}=\sqrt{9*89}=\sqrt{9}*\sqrt{89}=3\sqrt{89}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{89}}{2*-6}=\frac{3-3\sqrt{89}}{-12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{89}}{2*-6}=\frac{3+3\sqrt{89}}{-12} $
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