Answer :
It is quadratic equation.
First we have find delta given by formula: delta=[tex]b^{2}-4ac[/tex]
where our
a=16
b=-24
c=7
so, delta=[tex] 24^{2}-4*16*7=128[/tex]
Because delta is positive, there is real results.
Now we can use next formula x=[tex] \frac{-b+ \sqrt{delta} }{2a} [/tex]
, to find roots (results, 2 results because its quadratic equation and delta is greater than 0)
x1=[tex] \frac{24+ \sqrt{128} }{2*16} = \frac{3+ \sqrt{2} }{4} [/tex]
x2=[tex] \frac{24- \sqrt{128} }{2*16} = \frac{3- \sqrt{2} }{4} [/tex]
First we have find delta given by formula: delta=[tex]b^{2}-4ac[/tex]
where our
a=16
b=-24
c=7
so, delta=[tex] 24^{2}-4*16*7=128[/tex]
Because delta is positive, there is real results.
Now we can use next formula x=[tex] \frac{-b+ \sqrt{delta} }{2a} [/tex]
, to find roots (results, 2 results because its quadratic equation and delta is greater than 0)
x1=[tex] \frac{24+ \sqrt{128} }{2*16} = \frac{3+ \sqrt{2} }{4} [/tex]
x2=[tex] \frac{24- \sqrt{128} }{2*16} = \frac{3- \sqrt{2} }{4} [/tex]
[tex]16x^2-24x+7=0\\
16x^2-24x+9-2=0\\
(4x-3)^2=2\\
4x-3=\sqrt2 \vee 4x-3=-\sqrt2\\
4x=3+\sqrt2 \vee 4x=3-\sqrt2\\
x=\frac{3+\sqrt2}{4} \vee x=\frac{3-\sqrt2}{4}[/tex]
