2x^2+3x=275

Solution for 2x^2+3x=275 equation:

2x^2+3x=275

We move all terms to the left:

2x^2+3x-(275)=0

a = 2; b = 3; c = -275;
Δ = b2-4ac
Δ = 32-4·2·(-275)
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-47}{2*2}=\frac{-50}{4}
=-12+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+47}{2*2}=\frac{44}{4}
=11
$