4x^2+16x-6.5=0

Solution for 4x^2+16x-6.5=0 equation:

4x^2+16x-6.5=0

a = 4; b = 16; c = -6.5;
Δ = b2-4ac
Δ = 162-4·4·(-6.5)
Δ = 360
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:

$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-6\sqrt{10}}{2*4}=\frac{-16-6\sqrt{10}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+6\sqrt{10}}{2*4}=\frac{-16+6\sqrt{10}}{8}
$