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3x^2-3x=-2x+10
We move all terms to the left:
3x^2-3x-(-2x+10)=0
We get rid of parentheses
3x^2-3x+2x-10=0
We add all the numbers together, and all the variables
3x^2-1x-10=0
a = 3; b = -1; c = -10;
Δ = b2-4ac
Δ = -12-4·3·(-10)
Δ = 121
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-11}{2*3}=\frac{-10}{6} =-1+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+11}{2*3}=\frac{12}{6} =2 $
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