4.9x^2+10x=120

Solution for 4.9x^2+10x=120 equation:

4.9x^2+10x=120

We move all terms to the left:

4.9x^2+10x-(120)=0

a = 4.9; b = 10; c = -120;
Δ = b2-4ac
Δ = 102-4·4.9·(-120)
Δ = 2452
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:

$\sqrt{\Delta}=\sqrt{2452}=\sqrt{4*613}=\sqrt{4}*\sqrt{613}=2\sqrt{613}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{613}}{2*4.9}=\frac{-10-2\sqrt{613}}{9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{613}}{2*4.9}=\frac{-10+2\sqrt{613}}{9.8}
$