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3x^2+10=-11x
We move all terms to the left:
3x^2+10-(-11x)=0
We get rid of parentheses
3x^2+11x+10=0
a = 3; b = 11; c = +10;
Δ = b2-4ac
Δ = 112-4·3·10
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-1}{2*3}=\frac{-12}{6} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+1}{2*3}=\frac{-10}{6} =-1+2/3 $
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