1.5x+4.9x^2-5=0

Solution for 1.5x+4.9x^2-5=0 equation:

1.5x+4.9x^2-5=0

a = 4.9; b = 1.5; c = -5;
Δ = b2-4ac
Δ = 1.52-4·4.9·(-5)
Δ = 100.25
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{-(1.5)-\sqrt{100.25}}{2*4.9}=\frac{-1.5-\sqrt{100.25}}{9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.5)+\sqrt{100.25}}{2*4.9}=\frac{-1.5+\sqrt{100.25}}{9.8}
$