Step-by-step explanation:
[tex]-2x^2+10x=-14\qquad\text{divide both sides by (-2)}\\\\\dfrac{-2x^2}{-2}+\dfrac{10x}{-2}=\dfrac{-14}{-2}\\\\x^2-5x=7\qquad(a-b)^2=a^2-2ab+b^2\qquad(*)\\\\x^2-2(x)(2.5)=7\qquad\text{add}\ 2.5^2\ \text{to both sides}\\\\\underbrace{x^2-2(x)(2.5)+2.5^2}_{(*)}=7+2.5^2\\\\(x-2.5)^2=7+6.25\\\\(x-2.5)^2=13.25[/tex]
[tex]\text{If you want the solution, then:}\\\\(x-2.5)^2=13.25\iff x-2.5=\pm\sqrt{13.25}\\\\x-\dfrac{25}{10}=\pm\sqrt{\dfrac{1325}{100}}\\\\x-\dfrac{25}{10}=\pm\dfrac{\sqrt{1325}}{\sqrt{100}}\\\\x-\dfrac{25}{10}=\pm\dfrac{\sqrt{25\cdot53}}{10}\\\\x-\dfrac{25}{10}=\pm\dfrac{\sqrt{25}\cdot\sqrt{53}}{10}\\\\x-\dfrac{25}{10}=\pm\dfrac{5\sqrt{53}}{10}\\\\x-\dfrac{5}{2}=\pm\dfrac{\sqrt{53}}{2}\qquad\text{add}\ \dfrac{5}{2}\ \text{to both sides}\\\\x=\dfrac{5}{2}\pm\dfrac{\sqrt{53}}{2}[/tex]
[tex]\huge\boxed{x=\dfrac{5\pm\sqrt{53}}{2}}[/tex]
