4x^2=25x-21

Solution for 4x^2=25x-21 equation:

4x^2=25x-21

We move all terms to the left:

4x^2-(25x-21)=0

We get rid of parentheses

4x^2-25x+21=0

a = 4; b = -25; c = +21;
Δ = b2-4ac
Δ = -252-4·4·21
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-17}{2*4}=\frac{8}{8}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+17}{2*4}=\frac{42}{8}
=5+1/4
$