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3x^2+25x-5.3=0
a = 3; b = 25; c = -5.3;
Δ = b2-4ac
Δ = 252-4·3·(-5.3)
Δ = 688.6
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{-(25)-\sqrt{688.6}}{2*3}=\frac{-25-\sqrt{688.6}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{688.6}}{2*3}=\frac{-25+\sqrt{688.6}}{6} $
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