2x^2=17x-4.9=0

Solution for 2x^2=17x-4.9=0 equation:

2x^2=17x-4.9=0

We move all terms to the left:

2x^2-(17x-4.9)=0

We get rid of parentheses

2x^2-17x+4.9=0

a = 2; b = -17; c = +4.9;
Δ = b2-4ac
Δ = -172-4·2·4.9
Δ = 249.8
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{-(-17)-\sqrt{249.8}}{2*2}=\frac{17-\sqrt{249.8}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{249.8}}{2*2}=\frac{17+\sqrt{249.8}}{4}
$