Question

Решить квадратные уровнения: 1)3х²-4х-39=0 2)х²-5х=0 3)в²-5=0

Answer 1

$1) 3 x^{2} -4x-39=0\\D=16+468=484 \\ \sqrt{D} =22 \\ x_{1} = \frac{4+22}{6} = \frac{26}{6} = \frac{13}{3} =4 \frac{1}{3} \\ \\ x_{2} = \frac{4-22}{6} = \frac{-18}{6} =-3$

$2) x^{2} -5x=0 \\ x(x-5)=0\\x=0 \\ x-5=0\\x=5$

$3)b ^{2} -5=0\\b ^{2} =5\\b=+- \sqrt{5}$