20x^2+91x+49=0

Solution for 20x^2+91x+49=0 equation:

20x^2+91x+49=0

a = 20; b = 91; c = +49;
Δ = b2-4ac
Δ = 912-4·20·49
Δ = 4361
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:

$\sqrt{\Delta}=\sqrt{4361}=\sqrt{49*89}=\sqrt{49}*\sqrt{89}=7\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(91)-7\sqrt{89}}{2*20}=\frac{-91-7\sqrt{89}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(91)+7\sqrt{89}}{2*20}=\frac{-91+7\sqrt{89}}{40}
$