7x^2+49x-423=0

Solution for 7x^2+49x-423=0 equation:

7x^2+49x-423=0

a = 7; b = 49; c = -423;
Δ = b2-4ac
Δ = 492-4·7·(-423)
Δ = 14245
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-\sqrt{14245}}{2*7}=\frac{-49-\sqrt{14245}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+\sqrt{14245}}{2*7}=\frac{-49+\sqrt{14245}}{14}
$