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x2+4x-3=6x+12
We move all terms to the left:
x2+4x-3-(6x+12)=0
We add all the numbers together, and all the variables
x^2+4x-(6x+12)-3=0
We get rid of parentheses
x^2+4x-6x-12-3=0
We add all the numbers together, and all the variables
x^2-2x-15=0
a = 1; b = -2; c = -15;
Δ = b2-4ac
Δ = -22-4·1·(-15)
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*1}=\frac{-6}{2} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*1}=\frac{10}{2} =5 $
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