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5x^2+16x-48=0
a = 5; b = 16; c = -48;
Δ = b2-4ac
Δ = 162-4·5·(-48)
Δ = 1216
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{19}}{2*5}=\frac{-16-8\sqrt{19}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{19}}{2*5}=\frac{-16+8\sqrt{19}}{10} $
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