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5x^2+4x-1620=0
a = 5; b = 4; c = -1620;
Δ = b2-4ac
Δ = 42-4·5·(-1620)
Δ = 32416
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{32416}=\sqrt{16*2026}=\sqrt{16}*\sqrt{2026}=4\sqrt{2026}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{2026}}{2*5}=\frac{-4-4\sqrt{2026}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{2026}}{2*5}=\frac{-4+4\sqrt{2026}}{10} $
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