28x^2=-23x+15

Solution for 28x^2=-23x+15 equation:

28x^2=-23x+15

We move all terms to the left:

28x^2-(-23x+15)=0

We get rid of parentheses

28x^2+23x-15=0

a = 28; b = 23; c = -15;
Δ = b2-4ac
Δ = 232-4·28·(-15)
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{-(23)-47}{2*28}=\frac{-70}{56}
=-1+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+47}{2*28}=\frac{24}{56}
=3/7
$