4.9x^2+10x-140=0

Solution for 4.9x^2+10x-140=0 equation:

4.9x^2+10x-140=0

a = 4.9; b = 10; c = -140;
Δ = b2-4ac
Δ = 102-4·4.9·(-140)
Δ = 2844
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:

$\sqrt{\Delta}=\sqrt{2844}=\sqrt{36*79}=\sqrt{36}*\sqrt{79}=6\sqrt{79}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6\sqrt{79}}{2*4.9}=\frac{-10-6\sqrt{79}}{9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6\sqrt{79}}{2*4.9}=\frac{-10+6\sqrt{79}}{9.8}
$