Answer:
(a)
[tex]x=-4[/tex]
(b)
[tex]x=-\frac{5}{12}+i\frac{\sqrt{23}}{12},\:x=-\frac{5}{12}-i\frac{\sqrt{23}}{12}[/tex]
(c)
[tex]x=-\frac{1}{2},\:x=-\frac{3}{2}[/tex]
Step-by-step explanation:
(a)
[tex]-3x^2 - 24x -48=0[/tex]
we can compare it with standard quadratic equation
[tex]ax^2+bx+c=0[/tex]
we can find a, b and c
[tex]a=-3,b=-24,c=-48[/tex]
now, we can use quadratic formula
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
we can plug values
and we get
[tex]x=\frac{-\left(-24\right)\pm \sqrt{\left(-24\right)^2-4\left(-3\right)\left(-48\right)}}{2\left(-3\right)}[/tex]
and we get
[tex]x=-4[/tex]
(b)
[tex]6x^2+5x+2=0[/tex]
we can compare it with standard quadratic equation
[tex]ax^2+bx+c=0[/tex]
we can find a, b and c
[tex]a=6,b=5,c=2[/tex]
now, we can use quadratic formula
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
we can plug values
[tex]x=\frac{-5\pm \sqrt{5^2-4\cdot \:6\cdot \:2}}{2\cdot \:6}[/tex]
we get
[tex]x=-\frac{5}{12}+i\frac{\sqrt{23}}{12},\:x=-\frac{5}{12}-i\frac{\sqrt{23}}{12}[/tex]
(c)
[tex]4x^2+8x+3=0[/tex]
we can compare it with standard quadratic equation
[tex]ax^2+bx+c=0[/tex]
we can find a, b and c
[tex]a=4,b=8,c=3[/tex]
now, we can use quadratic formula
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
we can plug values
[tex]x=\frac{-8\pm \sqrt{8^2-4\cdot \:4\cdot \:3}}{2\cdot \:4}[/tex]
we get
[tex]x=-\frac{1}{2},\:x=-\frac{3}{2}[/tex]
