2x^2+21x-4.9=0

Solution for 2x^2+21x-4.9=0 equation:

2x^2+21x-4.9=0

a = 2; b = 21; c = -4.9;
Δ = b2-4ac
Δ = 212-4·2·(-4.9)
Δ = 480.2
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{-(21)-\sqrt{480.2}}{2*2}=\frac{-21-\sqrt{480.2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{480.2}}{2*2}=\frac{-21+\sqrt{480.2}}{4}
$