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(x)=3x^2+2x-5.
We move all terms to the left:
(x)-(3x^2+2x-5.)=0
We get rid of parentheses
-3x^2+x-2x+5.=0
We add all the numbers together, and all the variables
-3x^2-1x+5=0
a = -3; b = -1; c = +5;
Δ = b2-4ac
Δ = -12-4·(-3)·5
Δ = 61
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{61}}{2*-3}=\frac{1-\sqrt{61}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{61}}{2*-3}=\frac{1+\sqrt{61}}{-6} $
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