-(4x2+1)-(-3x2-6)=-3

Solution for -(4x2+1)-(-3x2-6)=-3 equation:

-(4x^2+1)-(-3x^2-6)=-3

We move all terms to the left:

-(4x^2+1)-(-3x^2-6)-(-3)=0

We add all the numbers together, and all the variables

-(-3x^2-6)-(4x^2+1)+3=0

We get rid of parentheses

3x^2-4x^2+6-1+3=0

We add all the numbers together, and all the variables

-1x^2+8=0

a = -1; b = 0; c = +8;
Δ = b2-4ac
Δ = 02-4·(-1)·8
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*-1}=\frac{0-4\sqrt{2}}{-2}
=-\frac{4\sqrt{2}}{-2}
=-\frac{2\sqrt{2}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*-1}=\frac{0+4\sqrt{2}}{-2}
=\frac{4\sqrt{2}}{-2}
=\frac{2\sqrt{2}}{-1}
$