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3x^2+10x-2.5=0
a = 3; b = 10; c = -2.5;
Δ = b2-4ac
Δ = 102-4·3·(-2.5)
Δ = 130
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{-(10)-\sqrt{130}}{2*3}=\frac{-10-\sqrt{130}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+\sqrt{130}}{2*3}=\frac{-10+\sqrt{130}}{6} $
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