Answer:
x = 1/2 or 3
Step-by-step explanation:
Compare the given equation to the standard form equation to identify the values of the variables in the formula. Then put those values in the formula and do the arithmetic.
[tex]ax^2+bx+c=0\\\\2x^2-7x+3=0\\\\a=2;\ b=-7;\ c=3\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x=\dfrac{-(-7)\pm\sqrt{(-7)^2-4(2)(3)}}{2(2)}=\dfrac{7\pm\sqrt{25}}{4}=\dfrac{7\pm 5}{4}\\\\x=\left\{\dfrac{1}{2},3\right\}[/tex]
Answer:
Step-by-step explanation:
[tex]2x^2-7x+3=0\\\\x_1 ,2 =\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=2\\\:b=-7\\,\:c=3\\:\quad x_{1,\:2}=\frac{-\left(-7\right)\pm \sqrt{\left(-7\right)^2-4\times\:2\times\:3}}{2\times\:2}\\\\=\frac{7+\sqrt{\left(-7\right)^2-4\times\:2\times\:3}}{2\times\:2}\\\\7+\sqrt{\left(-7\right)^2-4\times\:2\times\:3}=7+\sqrt{25}\\\\=\frac{7+\sqrt{25}}{2\times\:2}\\\\=\frac{7+\sqrt{25}}{4}\\\\\sqrt{25}=5\\\\=\frac{7+5}{4}\\\\=\frac{12}{4}\\\\=3 \\[/tex]
[tex]x=\frac{-\left(-7\right)-\sqrt{\left(-7\right)^2-4\times\:2\times\:3}}{2\times\:2}\\\\=\frac{7-\sqrt{\left(-7\right)^2-4\times\:2\cdot \:3}}{2\times\:2}\\\\7-\sqrt{\left(-7\right)^2-4\times \:2\times\:3}=7-\sqrt{25}\\\\=\frac{7-\sqrt{25}}{2\times \:2}\\\\=\frac{7-\sqrt{25}}{4}\\\sqrt{25}=5\\\\=\frac{7-5}{4}\\=\frac{2}{4}\\\\=\frac{1}{2}[/tex]
