4.9x^2+20x-51=0

Solution for 4.9x^2+20x-51=0 equation:

4.9x^2+20x-51=0

a = 4.9; b = 20; c = -51;
Δ = b2-4ac
Δ = 202-4·4.9·(-51)
Δ = 1399.6
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{-(20)-\sqrt{1399.6}}{2*4.9}=\frac{-20-\sqrt{1399.6}}{9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{1399.6}}{2*4.9}=\frac{-20+\sqrt{1399.6}}{9.8}
$