Решите уравнение.спасибо

$\sqrt{2x-3}+ \sqrt{4x+1}=4 2x-3 \geq 0, 4x+1 \geq 0 =\ \textgreater \ x \geq \frac{3}{2} ( \sqrt{2x-3} + \sqrt{4x+1})^{2} =16 2x-3+2 \sqrt{(2x-3)(4x+1)} +4x+1=16 2 \sqrt{(2x-3)(4x+1)} =16-6x+2=18-6x \sqrt{(2x-3)(4x+1)} =9-3x ( \sqrt{(2x-3)(4x+1)} )^{2} =(9-3x)^{2} (2x-3)(4x+1)=(9-3x)^{2} 8 x^{2} +2x-12x-3=81-54x+9 x^{2} x^{2} -44x+84=0 D = 44^{2} -4*84 = 1936-336=1600 x_{1} =(44-40)/2=4/2=2 x_{2}=(44+40)/2=84/2 =42$
Корень x₂ = 42 не подходит.

Ответ: x = 2.