21-5x^2=10x+4

Solution for 21-5x^2=10x+4 equation:

21-5x^2=10x+4

We move all terms to the left:

21-5x^2-(10x+4)=0

We get rid of parentheses

-5x^2-10x-4+21=0

We add all the numbers together, and all the variables

-5x^2-10x+17=0

a = -5; b = -10; c = +17;
Δ = b2-4ac
Δ = -102-4·(-5)·17
Δ = 440
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:

$\sqrt{\Delta}=\sqrt{440}=\sqrt{4*110}=\sqrt{4}*\sqrt{110}=2\sqrt{110}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{110}}{2*-5}=\frac{10-2\sqrt{110}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{110}}{2*-5}=\frac{10+2\sqrt{110}}{-10}
$