Question

Someone solve :
$y {}^{2} + 4y - 12$
​

Answer 1

Answer:

$\huge \pink{(y+6)(y-2)}$

Step-by-step explanation:

${y}^{2} + 4y - 12 =$

Sum:4=-6+(-2)

Product:-12=6×(-2)

$y{}^{2} 6y -2y - 12 = \\ y(y - 2) 6(y - 2) = \\ (y + 6)(y -2)$

Hope this helps☑️

$<marquee>Follow me if this helps u frnd....$

Answer 2

Solution :-

Given equation :-

→ y² + 4y - 12 = 0

Now by applying Sridhacharyas Formula.

$\dfrac{ -b \pm\sqrt{b^2 - 4ac}}{2a}$

$\implies y = \dfrac{ -4 \pm\sqrt{(4)^2 - 4(-12)(1)}}{2}$

$\implies y = \dfrac{ -4 \pm\sqrt{(16 +48}}{2}$

$\implies y = \dfrac{ -4 \pm\sqrt{64}}{2}$

$\implies y = \dfrac{ -4 \pm 8}{2}$

$\implies y = \dfrac{ -4 + 8}{2} \: or \dfrac{-4 - 8}{2}$

$\implies y = \dfrac{ 4}{2} \: or \dfrac{-12}{2}$

$\implies y = 2 \: or -6$