2x^2-224x+70=0

Solution for 2x^2-224x+70=0 equation:

2x^2-224x+70=0

a = 2; b = -224; c = +70;
Δ = b2-4ac
Δ = -2242-4·2·70
Δ = 49616
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:

$\sqrt{\Delta}=\sqrt{49616}=\sqrt{16*3101}=\sqrt{16}*\sqrt{3101}=4\sqrt{3101}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-224)-4\sqrt{3101}}{2*2}=\frac{224-4\sqrt{3101}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-224)+4\sqrt{3101}}{2*2}=\frac{224+4\sqrt{3101}}{4}
$