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x^2+((5x)/4)=61
We move all terms to the left:
x^2+((5x)/4)-(61)=0
We add all the numbers together, and all the variables
x^2+(+5x/4)-61=0
We get rid of parentheses
x^2+5x/4-61=0
We multiply all the terms by the denominator
x^2*4+5x-61*4=0
We add all the numbers together, and all the variables
x^2*4+5x-244=0
Wy multiply elements
4x^2+5x-244=0
a = 4; b = 5; c = -244;
Δ = b2-4ac
Δ = 52-4·4·(-244)
Δ = 3929
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{3929}}{2*4}=\frac{-5-\sqrt{3929}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{3929}}{2*4}=\frac{-5+\sqrt{3929}}{8} $
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