2x^2+176x+1445=0

Solution for 2x^2+176x+1445=0 equation:

2x^2+176x+1445=0

a = 2; b = 176; c = +1445;
Δ = b2-4ac
Δ = 1762-4·2·1445
Δ = 19416
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:

$\sqrt{\Delta}=\sqrt{19416}=\sqrt{4*4854}=\sqrt{4}*\sqrt{4854}=2\sqrt{4854}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(176)-2\sqrt{4854}}{2*2}=\frac{-176-2\sqrt{4854}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(176)+2\sqrt{4854}}{2*2}=\frac{-176+2\sqrt{4854}}{4}
$