(a) \(4x^{2} = 11x + 21 \implies 4x^{2} - 11x - 21 = 0\)
\(a = 4, b = - 11 , c = -21\)
\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
\(x = \frac{-(-11) \pm \sqrt{(-11)^{2} - 4(4)(-21)}}{2(4)}\)
\(x = \frac{11 \pm \sqrt{121 + 336}}{8}\)
\(x = \frac{11 \pm \sqrt{457}}{8}\)
\(x = \frac{11 \pm 21.378}{8}\)
\(x = \frac{11 + 21.378}{8}\) or \(x = \frac{11 - 21.378}{8}\)
\(x = \frac{32.378}{8}\) or \(x = \frac{-10.378}{8}\)\
\(x = \text{4.047 or -1.297}\)
\(x \approxeq \text{4.05 or -1.30}\).(2 decimal place).
(b) (i) \(I = \frac{PRT}{100}\)
\(I = £(1560 - 1500) = £60\)
\(60 = \frac{1500 \times R \times 1}{100}\)
\(60 = 15R \implies R = 4%\)
(ii) Interest for the second year
= \(\frac{1560 \times 4 \times 1}{100}\)
= \(£62.40\)
(c)

Initial price of the car	\(N300,000.00\)
Depreciation after first year (25%)	\(\frac{25}{100} \times 300,000 = N75,000\)
Value after 1st year	N225,000.00
Depreciation after 2nd year(20%)	\(\frac{20}{100} \times 225,000 = N45,000\)
Value after 2nd year	N180,000.00