-2.75+x^2+x=0

Solution for -2.75+x^2+x=0 equation:

-2.75+x^2+x=0

a = 1; b = 1; c = -2.75;
Δ = b2-4ac
Δ = 12-4·1·(-2.75)
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-2\sqrt{3}}{2*1}=\frac{-1-2\sqrt{3}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+2\sqrt{3}}{2*1}=\frac{-1+2\sqrt{3}}{2}
$