\(4x^2+3x-7\\ =\left(2x\right)^2+2.2x.\frac{3}{4}+\frac{9}{16}-\frac{121}{16}\\ =\left(2x+\frac{3}{4}\right)^2-\left(\frac{11}{4}\right)^2\\ =\left(2x-2\right)\left(2x+\frac{7}{2}\right)\\ =2\left(x-1\right)\left(2x+\frac{7}{2}\right)\)
4x2 + 3x - 7
= 4x2 - 4x + 7x - 7
= 4x(x - 1) + 7(x - 1)
= (x - 1)(4x + 7)