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3x^2+3x=-5x-5
We move all terms to the left:
3x^2+3x-(-5x-5)=0
We get rid of parentheses
3x^2+3x+5x+5=0
We add all the numbers together, and all the variables
3x^2+8x+5=0
a = 3; b = 8; c = +5;
Δ = b2-4ac
Δ = 82-4·3·5
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2}{2*3}=\frac{-10}{6} =-1+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2}{2*3}=\frac{-6}{6} =-1 $
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