Помогите решить уравнение:√2x+1=5-2x

$\sqrt{2x+1} =5-2x \\ 2x+1 \geq 0 \\ 2x \geq -1 \\ x \geq - \frac{1}{2} \\ x \geq -0.5$

x∈ $[-0.5;+ \infty )$

$\sqrt{2x+1} =5-2x \\ ( \sqrt{2x+1} ) ^{2} =(5-2x ) ^{2} \\ 2x+1=25+4 x^{2} -20x \\ 2x+1-25-4 x^{2} +20x=0 \\ -4 x^{2} +22x-24=0/*(-1)\\4 x^{2} -22x+24=0 /:2\\ 2 x^{2} -11x+12=0 \\ D=121-4*2*12=121-96=25 \\ \sqrt{D} =5 \\ x_{1} = \frac{11+5}{4} = \frac{16}{4} =4 \\ x_{2} = \frac{11-5}{4} = \frac{6}{4} =1.5$