Answer
The solutions are-$\frac{1}{2}$ or 4
Work Step by Step
x(2x-7)=4
2$x^{2}$-7x-4=0
2$x^{2}$-8x+x-4=0
2x(x-4)+1(x-4)=0
(2x+1)(x-4)=0
2x+1=0 or x-4 =0
x=-$\frac{1}{2}$ or x=4
The solutions are-$\frac{1}{2}$ or 4
Check
Let
x=-$\frac{1}{2}$
x(2x-7)=4
-$\frac{1}{2}$ [2(-$\frac{1}{2}$) -7]=4
-$\frac{1}{2}$ [-1 -7]=4
-$\frac{1}{2}$ [-8]=4
4=4
Let x=4
x(2x-7)=4
4(2$\times4$-7)=4
4(8-7)=1
4=4
