Chapter 9 - Quadratic Functions and Equations - 9-4 Factoring to Solve Quadratic Equations - Practice and Problem-Solving Exercises - Page 558: 28

10 ft by 25 ft

$x(2x+5) = 250$ $2x^2 +5x= 250$ $2x^2 +5x-250 = 250-250$ $2x^2 +5x-250 =0$ $2x^2-20x+25x-250 =0$ $2x(x-10)+25(x-10) =0$ $(2x+25)(x-10)=0$ $2x+25=0$ $2x+25-25 = 0-25$ $2x = -25$ $2x/2 = -25/2$ $x = -12.5$ $x-10=0$ $x-10+10 = 0+10$ $x = 10$ We can't have a negative length, so $x\ne-12.5$. $2x+5$ $2*10+5$ $20+5 =25$