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3x^2=+20x+12
We move all terms to the left:
3x^2-(+20x+12)=0
We add all the numbers together, and all the variables
3x^2-(20x+12)=0
We get rid of parentheses
3x^2-20x-12=0
a = 3; b = -20; c = -12;
Δ = b2-4ac
Δ = -202-4·3·(-12)
Δ = 544
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{34}}{2*3}=\frac{20-4\sqrt{34}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{34}}{2*3}=\frac{20+4\sqrt{34}}{6} $
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