Question

-(2x+3)^2+(x+5)(2x+5)=16

Answer 1

Объяснение:

–(2x+3)²+(x+5)(2x+5)=16

–(4х²+12х+9)+(2х²+5х+10х+25)=16

–4х²–12х–9+2х²+5х+10х+25=16

–2х²+3х+16=16

–2х²+3х=0

а=–2;

b=3

с=0

Д= 3²–4*(–2)*0= 9–0= 9

х(1)= $\frac{ - 3 + \sqrt{9} }{ - 4} = \frac{ - 3 + 3}{ - 4} = 0$

х(2)= $\frac{ - 3 - \sqrt{9} }{ - 4} = \frac{ - 3 - 3}{ - 4} = \frac{ - 6}{- 4} = \frac{3}{2} = 1.5$