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3x+10x^2=130
We move all terms to the left:
3x+10x^2-(130)=0
a = 10; b = 3; c = -130;
Δ = b2-4ac
Δ = 32-4·10·(-130)
Δ = 5209
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{5209}}{2*10}=\frac{-3-\sqrt{5209}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{5209}}{2*10}=\frac{-3+\sqrt{5209}}{20} $
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