# How do you find the solutions of 25c^2+20c=-8c+5c^2?

Solution: $c = 0 \mathmr{and} c = - \frac{7}{5}$
$25 {c}^{2} + 20 c = - 8 c + 5 {c}^{2} \mathmr{and} 25 {c}^{2} + 20 c + 8 c - 5 {c}^{2} = 0 \mathmr{and} 20 {c}^{2} + 28 c = 0 \mathmr{and} 4 c \left(5 c + 7\right) = 0 \mathmr{and} c \left(5 c + 7\right) = 0 \therefore c = 0 \mathmr{and} 5 c + 7 = 0 \mathmr{and} c = - \frac{7}{5}$
Solution: $c = 0 \mathmr{and} c = - \frac{7}{5}$ [Ans]