3.5x^2+16x-50=0

Solution for 3.5x^2+16x-50=0 equation:

3.5x^2+16x-50=0

a = 3.5; b = 16; c = -50;
Δ = b2-4ac
Δ = 162-4·3.5·(-50)
Δ = 956
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:

$\sqrt{\Delta}=\sqrt{956}=\sqrt{4*239}=\sqrt{4}*\sqrt{239}=2\sqrt{239}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{239}}{2*3.5}=\frac{-16-2\sqrt{239}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{239}}{2*3.5}=\frac{-16+2\sqrt{239}}{7}
$