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(3x)^2+15x-8.3=0
a = 3; b = 15; c = -8.3;
Δ = b2-4ac
Δ = 152-4·3·(-8.3)
Δ = 324.6
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{-(15)-\sqrt{324.6}}{2*3}=\frac{-15-\sqrt{324.6}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{324.6}}{2*3}=\frac{-15+\sqrt{324.6}}{6} $
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