Question

7xв квадрате +6x-4=0
Решите пожайлуста срочно

Answer 1

$\displaystyle\bf\\7x^{2} +6x-4=0\\\\D=6^{2} -4\cdot 7\cdot (-4)=36+112=148=(2\sqrt{37})^{2} \\\\\\x_{1} =\frac{-6-2\sqrt{37} }{14} =-\frac{3+\sqrt{37} }{7}\\\\\\x_{2} =\frac{-6+2\sqrt{37} }{14} =\frac{\sqrt{37} -3}{7} \\\\\\Otvet:-\frac{3+\sqrt{37} }{7} \ \ ; \ \ \frac{37-3}{7}$