0=4.9x^2-42x+35

Solution for 0=4.9x^2-42x+35 equation:

0=4.9x^2-42x+35

We move all terms to the left:

0-(4.9x^2-42x+35)=0

We add all the numbers together, and all the variables

-(4.9x^2-42x+35)=0

We get rid of parentheses

-4.9x^2+42x-35=0

a = -4.9; b = 42; c = -35;
Δ = b2-4ac
Δ = 422-4·(-4.9)·(-35)
Δ = 1078
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:

$\sqrt{\Delta}=\sqrt{1078}=\sqrt{49*22}=\sqrt{49}*\sqrt{22}=7\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-7\sqrt{22}}{2*-4.9}=\frac{-42-7\sqrt{22}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+7\sqrt{22}}{2*-4.9}=\frac{-42+7\sqrt{22}}{-9.8}
$