We have the equation:
[tex]x^2+8x-4=2x-7[/tex]We have to compare it to the equation:
[tex](x+s)^2+t=0[/tex]We start by rearranging the first equation:
[tex]\begin{gathered} x^2+8x-4=2x-7 \\ x^2+8x-2x-4+7=0 \\ x^2+6x+3=0 \end{gathered}[/tex]We then add a term to find the square in order to factorize it as the second equation:
[tex]\begin{gathered} x^2+6x+3=0 \\ x^2+2(3x)+3^2-3^2+3=0 \\ (x+3)^2-9+3=0 \\ (x+3)^2-6=0 \end{gathered}[/tex]Then, comparing to the second equation, s = 3 and t = -6.
Answer: the value of t for this quadratic equation is t = -6.
