Question

(х-7)(х+13)=0
(2,5-х)(х+4)=0​

Answer 1

Если это система то тогда так

Answer 2

(х-7)(х+13)=0

х² + 13х - 7х - 91 = 0

х² + 6х - 91 = 0

D = 6² - 4 • 1 • (-91) = 36 + 364 = 400

х1 =

$\frac{ - 6 + \sqrt{400} }{2 \times 1} = \frac{ - 6 + 20}{2} = \frac{14}{2 } = 7$

х2 =

$\frac{ - 6 - \sqrt{400} }{2 \times 1} = \frac{ - 6 - 20}{2} = \frac{ - 26}{2} = 13$

(2,5-х)(х+4)=0

2,5х + 10 - х² - 4х = 0

-х² - 1,5х + 10 = 0

D = (-1,5)² - 4 • (-1) • 10 = 2,25 + 40 = 42,25

х1 =

$\frac{ - ( - 1.5) + \sqrt{42.25} }{2 \times ( - 1)} = \frac{1.5 + 6.5}{ - 2} = \frac{8}{ - 2} = - 4$

х2 =

$\frac{ - ( - 1.5) - \sqrt{42.25} }{2 \times ( - 1)} = \frac{1.5 - 6.5}{2} = \frac{ - 5}{ - 2} = 2.5$