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5=3x^2+2x+3
We move all terms to the left:
5-(3x^2+2x+3)=0
We get rid of parentheses
-3x^2-2x-3+5=0
We add all the numbers together, and all the variables
-3x^2-2x+2=0
a = -3; b = -2; c = +2;
Δ = b2-4ac
Δ = -22-4·(-3)·2
Δ = 28
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{7}}{2*-3}=\frac{2-2\sqrt{7}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{7}}{2*-3}=\frac{2+2\sqrt{7}}{-6} $
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