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2x^2+-7x=6.125
We move all terms to the left:
2x^2+-7x-(6.125)=0
We add all the numbers together, and all the variables
2x^2-7x-6.125=0
a = 2; b = -7; c = -6.125;
Δ = b2-4ac
Δ = -72-4·2·(-6.125)
Δ = 98
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{98}=\sqrt{49*2}=\sqrt{49}*\sqrt{2}=7\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7\sqrt{2}}{2*2}=\frac{7-7\sqrt{2}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7\sqrt{2}}{2*2}=\frac{7+7\sqrt{2}}{4} $
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