Question

(4х+3)в квадрае = х в квадрате

Answer 1

$(4x+3) ^{2} = x^{2} \\ \\ 16 x^{2} +9+24x- x^{2} =0\\15 x^{2} +24x+9=0\\D=576-540=36 \\ \sqrt{D} =6 \\ x_{1} = \frac{-24-6}{30} = \frac{-30}{30} =-1 \\ \\ x_{2} = \frac{-24+6}{30} = \frac{-18}{30} =- \frac{6}{10} =-0.6$

Answer 2

(4x+3)^2=x^2
16x^2+24x+9=x^2
16x^2+24x+9-x^2=0
15x^2+24x+9=0 | :3
5x^2+8x+3=0
D=64-60=4
x1=(-8+2)/10=-6/10=-0.6
x2=(-8-2)/10=-10/10=-1