(4r^2+3r-1)-(3r^2-5r+4)=0

Solution for (4r^2+3r-1)-(3r^2-5r+4)=0 equation:

(4r^2+3r-1)-(3r^2-5r+4)=0

We get rid of parentheses

4r^2-3r^2+3r+5r-1-4=0

We add all the numbers together, and all the variables

r^2+8r-5=0

a = 1; b = 8; c = -5;
Δ = b2-4ac
Δ = 82-4·1·(-5)
Δ = 84
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{21}}{2*1}=\frac{-8-2\sqrt{21}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{21}}{2*1}=\frac{-8+2\sqrt{21}}{2}
$