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4x(2)+x-13=10
We move all terms to the left:
4x(2)+x-13-(10)=0
We add all the numbers together, and all the variables
4x^2+x-23=0
a = 4; b = 1; c = -23;
Δ = b2-4ac
Δ = 12-4·4·(-23)
Δ = 369
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{41}}{2*4}=\frac{-1-3\sqrt{41}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{41}}{2*4}=\frac{-1+3\sqrt{41}}{8} $
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