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x^2+3x=12-5x
We move all terms to the left:
x^2+3x-(12-5x)=0
We add all the numbers together, and all the variables
x^2+3x-(-5x+12)=0
We get rid of parentheses
x^2+3x+5x-12=0
We add all the numbers together, and all the variables
x^2+8x-12=0
a = 1; b = 8; c = -12;
Δ = b2-4ac
Δ = 82-4·1·(-12)
Δ = 112
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{7}}{2*1}=\frac{-8-4\sqrt{7}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{7}}{2*1}=\frac{-8+4\sqrt{7}}{2} $
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