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5x^2+3x+9=-10x+3
We move all terms to the left:
5x^2+3x+9-(-10x+3)=0
We get rid of parentheses
5x^2+3x+10x-3+9=0
We add all the numbers together, and all the variables
5x^2+13x+6=0
a = 5; b = 13; c = +6;
Δ = b2-4ac
Δ = 132-4·5·6
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-7}{2*5}=\frac{-20}{10} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+7}{2*5}=\frac{-6}{10} =-3/5 $
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