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5(x+3)=2x(x+3)
We move all terms to the left:
5(x+3)-(2x(x+3))=0
We multiply parentheses
5x-(2x(x+3))+15=0
We calculate terms in parentheses: -(2x(x+3)), so:We get rid of parentheses
2x(x+3)
We multiply parentheses
2x^2+6x
Back to the equation:
-(2x^2+6x)
-2x^2+5x-6x+15=0
We add all the numbers together, and all the variables
-2x^2-1x+15=0
a = -2; b = -1; c = +15;
Δ = b2-4ac
Δ = -12-4·(-2)·15
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-11}{2*-2}=\frac{-10}{-4} =2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+11}{2*-2}=\frac{12}{-4} =-3 $
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