Question

15y^2-41y+14=0 solve by factorization method ​

Answer 1

Step-by-step explanation:

$15y {}^{2} - 41y + 14 \\ \\ = 15y {}^{2} - 35y - 6y + 14 \\ \\ = 5y(3y - 7) - 2(3y - 7) \\ \\ = (5y - 2)(3y - 7)$

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Answer 2

Given:

Quadratic equation : 15y²- 41y+14=0

To Find:

Value of y

Solution:

it is given,

15y²- 41y+14=0    ax²+ bx+ c = 0

as a sign of c is -ve so you find factors of ac such that sum of factors is equal to b.

ac = 15 × 14 = 210 = 35 × 6

b = 32 + 5

so,

⇒ 15y²- 41y+14=0

⇒ 15y²- 35y - 6y + 14=0

⇒ 5y( 3y - 7) - 3( 2y - 7) =0

⇒ (5y - 3) ( 3y - 7) = 0

⇒ (5y - 3) = 0

y = 3/5

( 3y - 7) = 0

y = 7/3

Hence, the value y are 3/5 and 7/3.