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3x^2+10x=x^2+3x+15
We move all terms to the left:
3x^2+10x-(x^2+3x+15)=0
We get rid of parentheses
3x^2-x^2+10x-3x-15=0
We add all the numbers together, and all the variables
2x^2+7x-15=0
a = 2; b = 7; c = -15;
Δ = b2-4ac
Δ = 72-4·2·(-15)
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-13}{2*2}=\frac{-20}{4} =-5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13}{2*2}=\frac{6}{4} =1+1/2 $
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