7x^2+34x=48

Solution for 7x^2+34x=48 equation:

7x^2+34x=48

We move all terms to the left:

7x^2+34x-(48)=0

a = 7; b = 34; c = -48;
Δ = b2-4ac
Δ = 342-4·7·(-48)
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-50}{2*7}=\frac{-84}{14}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+50}{2*7}=\frac{16}{14}
=1+1/7
$