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0=6x^2+2x-99
We move all terms to the left:
0-(6x^2+2x-99)=0
We add all the numbers together, and all the variables
-(6x^2+2x-99)=0
We get rid of parentheses
-6x^2-2x+99=0
a = -6; b = -2; c = +99;
Δ = b2-4ac
Δ = -22-4·(-6)·99
Δ = 2380
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2380}=\sqrt{4*595}=\sqrt{4}*\sqrt{595}=2\sqrt{595}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{595}}{2*-6}=\frac{2-2\sqrt{595}}{-12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{595}}{2*-6}=\frac{2+2\sqrt{595}}{-12} $
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