5x^2-42x-49=0

Solution for 5x^2-42x-49=0 equation:

5x^2-42x-49=0

a = 5; b = -42; c = -49;
Δ = b2-4ac
Δ = -422-4·5·(-49)
Δ = 2744
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:

$\sqrt{\Delta}=\sqrt{2744}=\sqrt{196*14}=\sqrt{196}*\sqrt{14}=14\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-14\sqrt{14}}{2*5}=\frac{42-14\sqrt{14}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+14\sqrt{14}}{2*5}=\frac{42+14\sqrt{14}}{10}
$