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7x^2+10x-3=x^2+3x
We move all terms to the left:
7x^2+10x-3-(x^2+3x)=0
We get rid of parentheses
7x^2-x^2+10x-3x-3=0
We add all the numbers together, and all the variables
6x^2+7x-3=0
a = 6; b = 7; c = -3;
Δ = b2-4ac
Δ = 72-4·6·(-3)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-11}{2*6}=\frac{-18}{12} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+11}{2*6}=\frac{4}{12} =1/3 $
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