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5x^2=3x+12
We move all terms to the left:
5x^2-(3x+12)=0
We get rid of parentheses
5x^2-3x-12=0
a = 5; b = -3; c = -12;
Δ = b2-4ac
Δ = -32-4·5·(-12)
Δ = 249
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{249}}{2*5}=\frac{3-\sqrt{249}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{249}}{2*5}=\frac{3+\sqrt{249}}{10} $
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