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15x^2+25x-340=0
a = 15; b = 25; c = -340;
Δ = b2-4ac
Δ = 252-4·15·(-340)
Δ = 21025
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{21025}=145$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-145}{2*15}=\frac{-170}{30} =-5+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+145}{2*15}=\frac{120}{30} =4 $
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