21x=1.45x^2

Solution for 21x=1.45x^2 equation:

21x=1.45x^2

We move all terms to the left:

21x-(1.45x^2)=0

We get rid of parentheses

-1.45x^2+21x=0

a = -1.45; b = 21; c = 0;
Δ = b2-4ac
Δ = 212-4·(-1.45)·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-21}{2*-1.45}=\frac{-42}{-2.9}
=14+1/2.0714285714286
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+21}{2*-1.45}=\frac{0}{-2.9}
=0
$