2x2+21x-4.5=0

Solution for 2x2+21x-4.5=0 equation:

2x^2+21x-4.5=0

a = 2; b = 21; c = -4.5;
Δ = b2-4ac
Δ = 212-4·2·(-4.5)
Δ = 477
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:

$\sqrt{\Delta}=\sqrt{477}=\sqrt{9*53}=\sqrt{9}*\sqrt{53}=3\sqrt{53}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-3\sqrt{53}}{2*2}=\frac{-21-3\sqrt{53}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+3\sqrt{53}}{2*2}=\frac{-21+3\sqrt{53}}{4}
$