Question

find the zeroes of the quadratic polynomial
${y}^{2} + 7y - 60$
​

Answer 1

Step-by-step explanation:

${y}^{2} + 7y - 60 = 0\\ \\ {y}^{2} + 12y - 5y - 60 = 0 \\ \\ y(y + 12) - 5(y + 12) = 0 \\ \\ (y - 5)(y + 12) = 0 \\ \\ y = 5 \\ \\ y = - 12$

Hope it helps

Answer 2

Answer:-

y 2 +7y−60=0y 2 +12y−5y−60=0

y(y+12)−5(y+12)=0

(y−5)(y+12)=0

y=5

y=−12