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-5x^2+24x-3=0
a = -5; b = 24; c = -3;
Δ = b2-4ac
Δ = 242-4·(-5)·(-3)
Δ = 516
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{129}}{2*-5}=\frac{-24-2\sqrt{129}}{-10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{129}}{2*-5}=\frac{-24+2\sqrt{129}}{-10} $
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