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6x+3x^2=9x+5
We move all terms to the left:
6x+3x^2-(9x+5)=0
We get rid of parentheses
3x^2+6x-9x-5=0
We add all the numbers together, and all the variables
3x^2-3x-5=0
a = 3; b = -3; c = -5;
Δ = b2-4ac
Δ = -32-4·3·(-5)
Δ = 69
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{69}}{2*3}=\frac{3-\sqrt{69}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{69}}{2*3}=\frac{3+\sqrt{69}}{6} $
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