Question

$15y {}^{2} + 20y - 25$
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Answer 1

$\huge\tt\colorbox{white}{AnsweR}$

$\sf\:15y^{2}+20y-25$

Here, a = 15 , b = 20 and c = (-25)

Using Shridharacharya formula

$\tt\pink{x=\frac{-b± \sqrt{b^{2}-4ac}}{2a}}$

Now substituting the value of a, b and c

$\sf⟹\:y=\frac{-20± \sqrt{20^{2} - 4 \times 15 \times (-25)}}{2 \times 15}$

$\sf⟹\:y=\frac{-20 ± \sqrt {400 - 60 \times (-25)}}{30}$

$\sf⟹\:y=\frac{-20 ± \sqrt {400+1500}}{30}$

$\sf⟹\:y=\frac{-20 ± \sqrt { 1900}}{30}$

$\sf⟹\:y=\frac{-20± \sqrt{2 \times 2 \times 5 \times 5 \times 19}}{30}$

$\sf⟹\:y=\frac{-20± 2 \times 5 \sqrt {19}}{3}$

$\sf⟹\:y=\frac{-20± 10 \sqrt{ 19}}{3}$

$\sf\:Either\:y=\frac{-20+10 \sqrt{ 19}}{3}$

$\tt\pink{\:y=\frac{-10(2- \sqrt{ 19})}{3}}$

$\sf\:Either\:y=\frac{-20-10 \sqrt{ 19}}{3}$

$\tt\pink{\:y=\frac{-10(2+ \sqrt{ 19})}{3}}$

• Thnku ❣