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-16x^2+40x+1.5=0
a = -16; b = 40; c = +1.5;
Δ = b2-4ac
Δ = 402-4·(-16)·1.5
Δ = 1696
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1696}=\sqrt{16*106}=\sqrt{16}*\sqrt{106}=4\sqrt{106}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{106}}{2*-16}=\frac{-40-4\sqrt{106}}{-32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{106}}{2*-16}=\frac{-40+4\sqrt{106}}{-32} $
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