Processing Math: Done

From Förberedande kurs i matematik 1

The first step when we solve the second-degree equation is to complete the square on the left-hand side:

y2+2y−15=y+12−12−15=y+12−16 

The equation can now be written as

y+12=16 

and has, after taking the square root, the solutions

y+1=16=4  which gives y=−1+4=3

y+1=−16=−4  which gives y=−1−4=−5

A quick check shows that y=−5 and y=3 satisfy the equation:

y=−5 : LHS= −52+2−5−15=25−10−15=0  = RHS

y=3 : LHS= 32+23−15=9+6−15=0 = RHS