As it is a quadratic, first make it = 0.
2y^2 +10y +11 = 0 , " "ax^2 + bx + c =02y2+10y+11=0, ax2+bx+c=0
There are 3 options:
factorise - " " (this quadratic trinomial does not factorise)
completing the square (but a = 2, )
quadratic formula feels best for this one
y = (-b +-sqrt(b^2-4ac))/(2a) " with " a = 2, b= 10, c = 11y=−b±√b2−4ac2a with a=2,b=10,c=11
y = (-(10) +-sqrt(10^2-4(2)(11)))/(2(2))y=−(10)±√102−4(2)(11)2(2)
y = (-10 +-sqrt(100-88))/4y=−10±√100−884
y = (-10 +-sqrt12)/4y=−10±√124
We now solve it twice - once with + root and once with - root.
y = (-10 +sqrt12)/4 " or "y = (-10 -sqrt12)/4y=−10+√124 or y=−10−√124
y =-1.634 " or " y = -3.366y=−1.634 or y=−3.366
