(-3d^2+d-3)-(-5+3d)=0

Solution for (-3d^2+d-3)-(-5+3d)=0 equation:

(-3d^2+d-3)-(-5+3d)=0

We add all the numbers together, and all the variables

(-3d^2+d-3)-(3d-5)=0

We get rid of parentheses

-3d^2+d-3d-3+5=0

We add all the numbers together, and all the variables

-3d^2-2d+2=0

a = -3; b = -2; c = +2;
Δ = b2-4ac
Δ = -22-4·(-3)·2
Δ = 28
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{7}}{2*-3}=\frac{2-2\sqrt{7}}{-6}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{7}}{2*-3}=\frac{2+2\sqrt{7}}{-6}
$