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5x^2+25x-150=0
a = 5; b = 25; c = -150;
Δ = b2-4ac
Δ = 252-4·5·(-150)
Δ = 3625
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3625}=\sqrt{25*145}=\sqrt{25}*\sqrt{145}=5\sqrt{145}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5\sqrt{145}}{2*5}=\frac{-25-5\sqrt{145}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5\sqrt{145}}{2*5}=\frac{-25+5\sqrt{145}}{10} $
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