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3x^2=-9x-3
We move all terms to the left:
3x^2-(-9x-3)=0
We get rid of parentheses
3x^2+9x+3=0
a = 3; b = 9; c = +3;
Δ = b2-4ac
Δ = 92-4·3·3
Δ = 45
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{5}}{2*3}=\frac{-9-3\sqrt{5}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{5}}{2*3}=\frac{-9+3\sqrt{5}}{6} $
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