How do you solve 2y^2 + 34y +132 = 02y2+34y+132=0? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Mahek ☮ Apr 9, 2018 2y^2 + 34y +132 = 02y2+34y+132=0 => y^2 + 17y +66 = 0⇒y2+17y+66=0 => y^2 + 6y +11y+66 = 0⇒y2+6y+11y+66=0 => y(y + 6) +11(y+6) = 0⇒y(y+6)+11(y+6)=0 => (y+11)(y+6) = 0⇒(y+11)(y+6)=0 Therefore either y+11=0y+11=0 or y+6=0y+6=0 thats is y=-11y=−11 or y=-6y=−6 Answer link Related questions What are the different methods for solving quadratic equations? What would be the best method to solve -3x^2+12x+1=0−3x2+12x+1=0? How do you solve -4x^2+4x=9−4x2+4x=9? What are the two numbers if the product of two consecutive integers is 72? Which method do you use to solve the quadratic equation 81x^2+1=081x2+1=0? How do you solve -4x^2+4000x=0−4x2+4000x=0? How do you solve for x in x^2-6x+4=0x2−6x+4=0? How do you solve x^2-6x-16=0x2−6x−16=0 by factoring? How do you solve by factoring and using the principle of zero products x^2 + 7x + 6 = 0x2+7x+6=0? How do you solve x^2=2xx2=2x? See all questions in Comparing Methods for Solving Quadratics Impact of this question 2064 views around the world You can reuse this answer Creative Commons License