4.9x^2+1.05x=2

Solution for 4.9x^2+1.05x=2 equation:

4.9x^2+1.05x=2

We move all terms to the left:

4.9x^2+1.05x-(2)=0

a = 4.9; b = 1.05; c = -2;
Δ = b2-4ac
Δ = 1.052-4·4.9·(-2)
Δ = 40.3025
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{-(1.05)-\sqrt{40.3025}}{2*4.9}=\frac{-1.05-\sqrt{40.3025}}{9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.05)+\sqrt{40.3025}}{2*4.9}=\frac{-1.05+\sqrt{40.3025}}{9.8}
$