Question

решите уравнение (4х-3)²-16=0   

Answer 1

$(4x-3)^{2} -16=0$
(4x-3-4)(4x-3+4)=0
(4x-7)(4x+1)=0
$16 x^{2} -4x+28x-7=0$
$16 x^{2} -24x-7=0$
$x_{1,2}$=$frac{32+-32}{32} = left { {{0} atop {2}} right.$