(4x^2-3x-9)+(-9x+10)=0

Solution for (4x^2-3x-9)+(-9x+10)=0 equation:

(4x^2-3x-9)+(-9x+10)=0

We get rid of parentheses

4x^2-3x-9x-9+10=0

We add all the numbers together, and all the variables

4x^2-12x+1=0

a = 4; b = -12; c = +1;
Δ = b2-4ac
Δ = -122-4·4·1
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8\sqrt{2}}{2*4}=\frac{12-8\sqrt{2}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8\sqrt{2}}{2*4}=\frac{12+8\sqrt{2}}{8}
$