Question

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а) 2$x^{2}$ + 10х +12 =0
б) -3$x^{2}$ + 18х -24 =0
в) 0.6$x^{2}$ + 0.8х - 7.8 =0

Answer 1

$2 x^{2} +10x+12=0 \\ x^{2} +5x+6=0 \\ a=1;b=5;c=6 \\ D= b^{2}-4ac=25-4*6=25-24=1 \\ \\ x_{1} = \frac{-b+ \sqrt{D} }{2a} = \frac{-5+1}{2} = \frac{-4}{2} =-2 \\ \\ x_{2} =\frac{-b- \sqrt{D} }{2a}= \frac{-5-1}{2} = \frac{-6}{2} =-3$

$-3 x^{2} +18x-24=0 \\ x^{2} -6x+8=0 \\ a=1;b=-6;c=8 \\ D= b^{2} -4ac=36-4*8=36-32=4 \\ \\ x_{1} = \frac{-b+ \sqrt{D} }{2a} = \frac{6+2}{2} = \frac{8}{2} =4 \\ \\ x_{1} = \frac{-b- \sqrt{D} }{2a} = \frac{6-2}{2} = \frac{4}{2} =2$


$0,6 x^{2} +0,8x-7,8=0 \\ 6 x^{2} +8x-78=0 \\ 3 x^{2} +4x-39=0 \\ a=3;b=4;c=-39 \\ D= b^{2} -4ac=16+4*3*39=16+468=484 \\ x_{1} = \frac{-b+ \sqrt{D}}{2a} = \frac{-4+22}{2*3} = \frac{18}{6} =3 \\ x_{2} = \frac{-b- \sqrt{D}}{2a} = \frac{-4-22}{2*3} = \frac{26}{6} = \frac{13}{3} =4 \frac{1}{3}$