x^2-49x+13=0

Solution for x^2-49x+13=0 equation:

x^2-49x+13=0

a = 1; b = -49; c = +13;
Δ = b2-4ac
Δ = -492-4·1·13
Δ = 2349
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:

$\sqrt{\Delta}=\sqrt{2349}=\sqrt{81*29}=\sqrt{81}*\sqrt{29}=9\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-49)-9\sqrt{29}}{2*1}=\frac{49-9\sqrt{29}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+9\sqrt{29}}{2*1}=\frac{49+9\sqrt{29}}{2}
$