49x^2-212x+120=0

Solution for 49x^2-212x+120=0 equation:

49x^2-212x+120=0

a = 49; b = -212; c = +120;
Δ = b2-4ac
Δ = -2122-4·49·120
Δ = 21424
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:

$\sqrt{\Delta}=\sqrt{21424}=\sqrt{16*1339}=\sqrt{16}*\sqrt{1339}=4\sqrt{1339}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-212)-4\sqrt{1339}}{2*49}=\frac{212-4\sqrt{1339}}{98}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-212)+4\sqrt{1339}}{2*49}=\frac{212+4\sqrt{1339}}{98}
$