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2x^2+16x-3=0
a = 2; b = 16; c = -3;
Δ = b2-4ac
Δ = 162-4·2·(-3)
Δ = 280
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{70}}{2*2}=\frac{-16-2\sqrt{70}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{70}}{2*2}=\frac{-16+2\sqrt{70}}{4} $
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