0=10+42x+4.9x^2

Solution for 0=10+42x+4.9x^2 equation:

0=10+42x+4.9x^2

We move all terms to the left:

0-(10+42x+4.9x^2)=0

We add all the numbers together, and all the variables

-(10+42x+4.9x^2)=0

We get rid of parentheses

-4.9x^2-42x-10=0

a = -4.9; b = -42; c = -10;
Δ = b2-4ac
Δ = -422-4·(-4.9)·(-10)
Δ = 1568
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:

$\sqrt{\Delta}=\sqrt{1568}=\sqrt{784*2}=\sqrt{784}*\sqrt{2}=28\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-28\sqrt{2}}{2*-4.9}=\frac{42-28\sqrt{2}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+28\sqrt{2}}{2*-4.9}=\frac{42+28\sqrt{2}}{-9.8}
$