Given the equation
[tex](6x^2-7x+2)(x^2-5x+5)\text{ = 0}[/tex]
The equations in bracket are both quadratic equations, hence we are going to equate both to zero and factorize as shown;
[tex]\begin{gathered} 6x^2-7x+2=0 \\ x^2-5x+5\text{ =0} \end{gathered}[/tex]
Factorize both equations;
For 6x^2-7x+2 = 0
6x^2-3x-4x+2 = 0
3x(2x-1)-2(2x-1) = 0
(3x-2)(2x-1) = 0
3x-2 = 0 and 2x-1 = 0
3x = 2 and 2x = 1
x = 2/3 and x = 1/2
Also for the second equation;
x^2-5x+5 = 0
Using the general formula;
[tex]\begin{gathered} x=\frac{-(-5)\pm\sqrt[]{(-5)^2-4(1)(5)}}{2} \\ x\text{ =}\frac{5\pm\sqrt[]{25^{}-20}}{2} \\ x\text{ = }\frac{5\pm\sqrt[]{5}}{2} \\ x\text{ = }\frac{5+2.24}{2}\text{ and }\frac{5-2.24}{2} \\ x=\frac{7.24}{2}\text{ and }\frac{2.74}{2} \\ x\text{ = 3.62 and 1}.37 \end{gathered}[/tex]
Hence the values of x are 2/3, 1/2, 3.62 and 1.37
